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\topmatter \title On lower semicontinuity\\and relaxation\endtitle
\author Irene Fonseca\,\dag\quad\&\quad Giovanni
Leoni\,\ddag\endauthor \abstract Lower semicontinuity and relaxation
results in BV are obtained for multiple integrals
$$ F(u,\Omega):=\int_\Omega f(x,u(x),\nabla u(x))\,dx, \quad u \in
W^{1,1}(\Omega;\Bbb R^d),
$$
where the energy density $f$ satisfies lower semicontinuity conditions with respect to $(x,u)$ and is not subjected to coercivity hypotheses. These results call for methods recently developed in the Calculus of Variations.                 
\endabstract



\subjclass 49J45,  49Q20\endsubjclass

\keywords Functions of bounded variation, lower semicontinuity,
relaxation, convexity, quasiconvexity\endkeywords



\endtopmatter
\leftheadtext\nofrills{I. Fonseca and G. Leoni}
\rightheadtext\nofrills{Lower semicontinuity and relaxation}
\document
\footnote""{\dag\quad Department of Mathematical Sciences, Carnegie Mellon
University, Pittsburgh, PA 15213}
\footnote""{\ddag\quad   Dipartimento di Scienze e Tecnologie
Avanzate, Universit\'a del Piemonte Orientale, Alessandria, Italy 15100.}
\noindent {\bf\S1. Introduction.}
\medskip

In this paper we address lower semicontinuity and relaxation properties for multiple functionals of the form
$$F(u,\Omega):=\int_\Omega f(x,u(x),\nabla u(x))\,dx,
$$ 
 where $\Omega$ is an open  subset of $\Bbb R^N$, and $u(x)$ is
a $\Bbb R^d$--valued function defined on $\Omega$.

In \cite{30} Serrin considered the scalar--valued case where
$$f\in C(\Omega\times\Bbb R\times \Bbb R^N;[0,\infty))\quad \text{and}
\quad f(x,u,\cdot)\quad\text{ is convex in $\Bbb R^N$.}\tag1.1$$
Among his results we select Theorems A and B below.
\medskip
\noindent\proclaim{Theorem A} {\rm (cf. \cite{30, Theorem 11})} Assume that
$f$ satisfies $(1.1)$. Let $u\in BV\loc(\Omega;\Bbb R)$, and let
$\{u_n\}$ be a sequence of functions in $ W^{1,1}\loc(\Omega;\Bbb R)$
 converging to $u$ in $L^1\loc(\Omega;\Bbb R)$. Let $\lambda$,
 $\rho$ be moduli of continuity such that
\smallskip
\item{\rm (i)}  $\rho(s)\le C\, s$ for $C>0$ and all $s>0$ large, and
$$ |f(x,u,\xi)-f(x_0,u_0,\xi)|\le\lambda(|x-x_0|)(1+f(x,u,\xi))
+  \rho(|u-u_0|)$$
for all $(x,u)$,
$(x_0,u_0)\in \Omega\times\Bbb R,$ and for all
$\xi\in\Bbb R^ N$;

\noindent
or
\item{\rm (ii)} $ \dsize |f(x,u,\xi)-f(x_0,u_0,\xi)|\le
\lambda(|x-x_0|+|u-u_0|)(1+f(x,u,\xi))$  for all $(x,u)$,
$(x_0,u_0)\in \Omega\times\Bbb R,$ and for all
$\xi\in\Bbb R^ N$. Suppose, in addition, that
  $u(x)$ is continuous.

\noindent Then
$$\int_\Omega f(x,u(x),\nabla u(x))\,dx
\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx.$$
\endproclaim
\medskip
Here $\nabla u$ is the Radon--Nikodym derivative of
 the distributional derivative $Du$ of $u$, with respect to the
$N$--dimensional Lebesgue measure $\Cal L^N$. Also, we intend by {\it modulus of continuity} a nonnegative, increasing, continuous function $\rho$ such that $\rho(0) = 0.$
\par\

\noindent\proclaim{Theorem B} {\rm (cf. \cite{30, Theorem 12})} Assume that
$f$ satisfies $(1.1)$ and any one of the following conditions:\smallskip
\item{\rm (i)\phantom{ii}} $f(x,u,\xi)\to\infty$ as $|\xi|\to\infty$
for each $(x,u)\in\Omega\times\Bbb R$.
\item{\rm (ii)\phantom{i}} $f(x,u,\cdot)$ is strictly convex in $\Bbb R^N$
for each $(x,u)\in\Omega\times\Bbb R$.
\item{\rm (iii)} The derivatives $f_x$, $f_\xi$ and $f_{\xi\,x}$ exist and are
continuous.
\smallskip\noindent
Then
$F(u,\Omega)$ is lower semicontinuous in
$ W^{1,1}\loc(\Omega;\Bbb R)$ with respect to local convergence in $L^1$.
\endproclaim
\medskip
The prototype of integrands that we want to study is represented by $f=h(x,u)\,|\xi|$, where $h\ge 0$, for which conditions
 (i)--(iii) of Theorem B may be violated; hence, in this paper we will focus our attention mainly
 on Theorem A. Note also that while
conditions (i) and (ii) of Theorem A are trivially satisfied when $f=f(\xi)$, so that
$L^1\loc$ lower semicontinuity holds in this case  only under assumption
  (1.1),  Theorem B is more stringent as it imposes extra conditions on the
 dependence of $f$ on the gradient variable $\xi$.
\par It is worth noticing that Theorem A requires no {\it coercivity hypothesis}, i.e., a condition of the type
$$
f(x,u,\xi) \geq C|\xi| - \frac{1}{C}
$$
for some constant $C>0$.
 One of the main  purposes of this paper is to try to understand the
deep relation between  coercivity (or the lack of it) and lower semicontinuity.
A drawback in Theorem A(ii) is that,
in practice, one seldom knows whether the target function
$u(x)$ is continuous or not.
Important examples of integrands which  satisfy (i) and (ii) of
Theorem A are given by
$$f=f(x,\xi)=h(x)g(\xi),\qquad f=f(x,u,\xi)=h(x,u)g(\xi),$$
where $h(x)$ and $h(x,u)$ are uniformly continuous functions bounded away from
zero
 and $g$ is a nonnegative convex function.
Conditions (i) and (ii) of Theorem A appear often in the study
of lower semicontinuity and relaxation and were exploited by several authors.
Dal Maso \cite{9} obtained
an integral representation formula for
the relaxation $\Cal F(u,\Omega)$
with respect to the
$L^1\loc$ topology of the functional $F$, namely
$$\Cal F(u,\Omega):=\inf_{\{u_n\}}\left\{\liminf_{n\to\infty}F(u_n,\Omega):\, u_n
\in W^{1,1}\loc(\Omega;\Bbb R),\, u_n\to u\quad\text{in}\quad
 L^1\loc(\Omega;\Bbb R)\right\},$$
under a weak form of  condition (ii) in Theorem A  and assuming coercivity. Let
$$H(u,\Omega):=\int_\Omega f(x,u,\nabla u)\,dx +\int_\Omega
f^\infty(x,u,dC(u))
+\int_{S(u)\cap\Omega}\left(
\int_{u^-(x)}^{u^+(x)}f^\infty(x,s,\nu_u)\,ds\right)
d\Cal H^{N-1}(x),\tag1.2$$
 where $f^\infty$ is the {\it
recession function}
 of $f$, that is $\dsize f^\infty(x,u,\xi):
 =\limsup_{t\to\infty}\frac{f(x,u,t\,\xi)}t$,
 $C(u)$ is the Cantor
part of $Du$, and
$(u^+-u^-)$ is the jump of $u$ across the interface $S(u)$.\par
As a corollary of
Dal Maso's  general results\footnote{Dal Maso's results are given in terms
of $\Gamma$--convergence of a family of functionals
$F_n(u,\Omega):=\int_\Omega f_n(x,u(x),\nabla u(x))\,dx$. We take here
$f_h\equiv f$ and refer to \cite{9} for the more general statement
of Theorems 3.2 and 3.5.}, we have the following theorem
\noindent\proclaim{Theorem C} {\rm  (cf. \cite{9, Theorem 3.2})} Assume that
$\Omega$ is bounded and that
$f$ is a Borel function which satisfies $(1.1)$
for $\Cal H^N$ a.e $(x,u) \in \Omega\times\Bbb R$.
Suppose also that
 for every $r>0$ there exist $C>0$ and three functions $c$, $a$ and $A\in
C(\Omega;[0,\infty))\cap L^1(\Omega;[0,\infty))$, with $c(x)>0$ in $\Omega$,
 such that
$$ c(x)|\xi|-a(x)\le f(x,u,\xi)\le C|\xi|+A(x)\tag1.3$$
 for all $(x,u,\xi) \in \Omega\times\Bbb R\times\Bbb R^N$
 with $|u|\le r$.
Finally, assume  that
 for $\Cal H^N$ a.e $(x_0,u_0) \in \Omega\times\Bbb R$
and  for every  $\varepsilon>0$ there exists
$\delta>0$ such that
$$ |f(x,u,\xi)-f(x_0,u_0,\xi)|\le\varepsilon(1+|\xi|)
\tag1.4
$$  for all $(x,u) \in \Omega\times\Bbb R$
 with $|x-x_0|+|u-u_0|\le\delta$ and for all
$\xi\in\Bbb R^ N$. Then for all
$u\in BV(\Omega;\Bbb R)\cap L^\infty(\Omega;\Bbb R)$ we have
$$\aligned
\Cal F(u,\Omega)=H(u,\Omega).
\endaligned\tag1.5
$$
Furthermore, if there exist $b\in L^1(\Omega;[0,\infty))$, $C\ge 0$,
$\alpha\ge 1$, such that
$f(x,u,0)\le C|u|^\alpha+b(x)$
 for all $(x,u) \in \Omega\times\Bbb R$, then
$(1.5)$ holds for all  $u\in BV(\Omega;\Bbb R)\cap L^\alpha(\Omega;\Bbb R)$.
\endproclaim
\medskip
Note that (1.5) implies, in particular, that $F$ is $L^1$-lower semicontinuous in $W^{1,1}$. Indeed, as $f^\infty\ge 0$ it follows that
$\int_\Omega f(x,u(x),\nabla u(x))\,dx\le \Cal F(u,\Omega)$ for $u \in BV(\Omega;\Bbb R)$. Therefore, Dal Maso's result extends
Theorem A(ii) of Serrin to target functions $u$ of bounded variation {\it  which are not necessarily
 continuous}. However, the ``price'' to pay for this extension is the
 {\it coercivity and growth assumption}
(1.3). Dal Maso established  also  an integral representation
result for functionals which are not necessarily coercive.
\noindent\proclaim{Theorem D} {\rm  (cf. \cite{9, Theorem 3.5})} Assume that
$\Omega$ is bounded,  that
$f$ is a Borel function, and that $f(x,u,\cdot)$ is positively homogeneous
 of degree one and convex.  Suppose also that  there exist
$\lambda:\Omega\times\Bbb R\to[0,\infty)$, with $\lambda(\cdot,u)$ continuous
and $\lambda(0,u)=0$ for $\Cal L^1$ a.e. $u\in\Bbb R$, and a function
$P\in C(\Omega;[0,\infty))$
 such that
$$ |f(x,u,\xi)-f(x_0,u,\xi)|\le\lambda(|x-x_0|,u)(1+f(x,u,\xi)),\qquad
0\le f(x,u,\xi)\le P(u)|\xi|,
\tag1.6
$$  for all $x$, $x_0 \in \Omega$, $u\in\Bbb R$
 and  $\xi\in\Bbb R^ N$.
Then $(1.5)$ holds for all $u\in BV(\Omega;\Bbb R)$.
\endproclaim
\medskip
The main difference between hypotheses (1.4) and (1.6)$_1$ is that (1.4) is
a {\it local} hypothesis in $(x,u)$, while $(1.6)_1$ may be
interpreted as a global restriction in $u$.
The conditions of Theorem D are satisfied by
$$f(x,u,\xi)=h(x)B(u)|\xi|$$
where $h$ is a positive, bounded, uniformly continuous function   and
 $B$ is a nonnegative continuous function. Note also that condition (1.6)$_1$ is trivially
 satisfied when $f$ does not depend on $x$, that is
$f=f(u,\xi)$.
 Lower semicontinuity for these  integrands of the form $f=f(u,\xi)$
 was later studied by
De Giorgi, Buttazzo and  Dal Maso
 \cite{13}, who proved the following result:

\noindent\proclaim{Theorem E} {\rm  (cf. \cite{13, Theorem 1})} Assume that
$f=f(u,\xi)$ is nonnegative,
 measurable in the variable $u$, and convex in $\xi$.
 Suppose also that $f(u,0)$ is lower semicontinuous and that
$$\limsup_{\xi\to0}\frac{(f(u,0)-f(u,\xi))^+}{|\xi|}
\in L^1\loc(\Bbb R;\Bbb R).$$
\par
Then for every  $u\in  W^{1,1}\loc(\Omega;\Bbb R)$ the function
$f(u(x),\nabla u(x))$ is measurable and the functional
$F(u,\Omega)$ is lower semicontinuous in
$ W^{1,1}\loc(\Omega;\Bbb R)$ with respect to local convergence in $L^1$.
\endproclaim
\medskip
 Theorem E was  extended by De Cicco
 in \cite{12} to functions of bounded variation. More precisely, De Cicco showed
that when $f=f(u,\xi)$ satisfies the hypotheses of Theorem E then the functional
$H(u)$ defined in (1.2)  is lower semicontinuous in
$ BV\loc(\Omega;\Bbb R)$ with respect to local convergence in $L^1$.
The hypotheses of Theorem E
were significantly weakened by Ambrosio in  \cite{3} (see also
\cite{11}), where the sequence
$\{u_n\}$ is assumed to be bounded
  in $ W^{1,1}(\Omega;\Bbb R)$.
  This condition is somewhat related
to coercivity, and  we will not dwell more on it here.  \smallskip
Unlike the case where $f=f(u,\xi)$,  {\it without some kind of
coercivity}
one cannot expect in general  lower semicontinuity
in the $L^1$ topology for functionals of the form
$$\int_\Omega f(x,\nabla u(x))\,dx.$$
Indeed,  in \cite{9} Dal Maso, following a
counterexample of Aronszajn,  constructed a continuous function
$\omega:\Omega\to \Bbb R$, where $\Omega=(0,1)\times (0,1)$ and
$x=(x_1,x_2)$, and a sequence of functions $\{u_n\}$
 converging to $u(x)=x_2$ in $L^\infty(\Omega;\Bbb R)$, such that
$$\int_\Omega |(\sin \omega (x),\cos \omega(x))\cdot\nabla u(x))|\,dx
>\liminf_{n\to\infty}
\int_\Omega |(\sin \omega(x),\cos \omega(x))\cdot\nabla u_n(x))|\,dx.$$
Since the target
function $u(x_1,x_2)=x_2$ is continuous,
$f(x,\xi)=|(\sin \omega (x),\cos \omega(x))\cdot\xi|$
cannot satisfy  either condition (i) or (ii) of Theorem A, for in this case we would obtain a
contradiction. This example  suggests that, when there is no coercivity,
lower semicontinuity
in the $L^1$ topology may fail unless
 we strengthen  (1.1) with an uniform continuity condition.\smallskip
We are now ready to present the main result of the paper.

\medskip
\noindent\proclaim{Theorem 1.1}  Assume that
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ is a  Borel integrand,
$f(x,u,\cdot)$ is convex in $\Bbb R^N$, and  for all $(x_0,u_0)\in \Omega\times\Bbb R$ and $\varepsilon>0$ there
exists $\delta>0$ such that
$$f(x_0,u_0,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))
\tag1.7$$
for  all $(x,u)\in\Omega\times\Bbb R$
 with $|x-x_0|+|u-u_0|\le\delta$ and for all
$\xi\in\Bbb R^ N$.
 Let   $u\in BV\loc(\Omega;\Bbb R)$, and let
$\{u_n\}$ be a sequence of functions in $ W^{1,1}\loc(\Omega;\Bbb R)$
  converging to $u$ in $L^1\loc(\Omega;\Bbb R)$. Then
$$H(u,\Omega)\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx.
$$
\endproclaim
\medskip
The main tool in the proof of Theorem 1.1 is the {\it blow--up method}
 introduced by Fonseca and M\"uller \cite{17}, \cite{18}, where
  we reduce the domain $\Omega$ to a ball and the target function $u$ becomes a piecewise affine function.
Since affine functions are locally bounded, in the scalar case we
 may replace the truncation used in \cite{17},
\cite{18}, in a vectorial setting, and which required
a degenerate coercivity condition, by  a considerably simpler argument.

\smallskip
Theorem 1.1 improves Serrin's Theorem A, not only because  continuity
of the target function
$u$ is assumed in Theorem A(ii), and is not needed here, but also
because condition (1.7) is
significantly weaker than  (ii), as the following result
illustrates:

\medskip
\noindent\proclaim{Corollary 1.2} Let $g:\Bbb R^N\to[0,\infty)$ be
a convex function, and let $h:\Omega\times\Bbb R\to[0,\infty)$
be a lower semicontinuous  function. If $u\in BV\loc(\Omega;\Bbb R)$
and
$\{u_n\} \subset W^{1,1}\loc(\Omega;\Bbb R)$
 converges to $u$ in $L^1\loc(\Omega;\Bbb R)$, then
$$\int_\Omega h(x,u)g(\nabla u)\,dx
\le\liminf_{n\to\infty}
\int_\Omega h(x,u_n)g(\nabla u_n)\,dx.$$
\endproclaim
\medskip
This result seems to be new in this generality.
Note that conditions (1.3) and (1.4) in Theorem C of Dal Maso
imply the validity of (1.7), while
 $f(x,u,\xi)=h(x,u)g(\xi)$ as in Corollary 1.2 satisfies (1.7), but not, in general,
(1.4), (1.6)$_1$, and (i), (ii) of Theorem A.

\par
Conditions of the type (1.7) appeared already in the papers
of Fonseca and M\"uller \cite{17}, \cite{18},
Dal Maso and Sbordone \cite{10}, Fusco and Hutchinson  \cite{20}.
 All these results deal with the vectorial case and require some type of
coercivity conditions. \par
 In the special case where
 $$h=h(x):=\chi_A(x)\qquad\text{for some measurable set
$A\subset \Omega$,}$$
then $h(x)g(\xi)$ satisfies (1.7) if and only if
$\Cal L^N(\partial A)=0$ (i.e. if $\chi_A(x)$ has a lower semicontinuous
representative),
and thus we recover the condition obtained by Gangbo
 \cite{21}.
Corollary 1.2 attests to the  sharpness of  condition  (1.7). Indeed,  when $N=1$ and
$\Omega$ is bounded,
 Fusco \cite{19}  proved that the functional
$$F(u):=\int_\Omega h(x)|u'(x)|\, dx,\qquad u\in  W^{1,1}(\Omega;\Bbb R),$$
where $h(x)$ is a bounded, nonnegative measurable function,
 is lower semicontinuous in $L^1(\Omega;\Bbb R)$ if and only if
$h(x)$ is lower semicontinuous.
\medskip
\noindent\proclaim{Theorem 1.3} Assume that $\Omega$ is bounded,
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ is a Borel integrand,
$f(x,u,\cdot)$ is  convex in $\Bbb R^N$, and
there exists a constant $C>0$
 such that
$$ 0\le f(x,u,\xi)\le C(1+|\xi|)\qquad\text{
 for all $(x,u,\xi) \in \Omega\times\Bbb R\times\Bbb R^N$. }
 \tag1.8$$
 Then $\Cal F(u,\cdot)$ is the trace of a finite, Radon measure on the
 open subsets of $\Omega$, and \smallskip
\item{\rm (i)} if $f$ is Carath\'eodory or
$f(\cdot,\cdot,\xi)$ is upper semicontinuous then
$\dsize\Cal F(u,\Omega\backslash(S(u)\cup M(u)))\le
\int_\Omega f(x,u,\nabla u)\,dx;$
\item{}
\item{\rm (ii)} if $f^\infty(\cdot,\cdot,\xi)$ is upper semicontinuous then
$\dsize\Cal F(u,M(u))\le \int_\Omega f^\infty(x,u,dC(u));$
\item{}
\item{\rm (iii)} if $f^\infty(\cdot,u,\xi)$ is upper semicontinuous then
$\dsize\Cal F(u,S(u))\le \int_{S(u)\cap\Omega}\left(
\int_{u^-(x)}^{u^+(x)}f^\infty(x,s,\nu_u)\,ds\right)d\Cal H^{N-1}(x).$
\endproclaim
\medskip
Here, and in what follows, $M(u)$ is the support of the Cantor part of $Du$.
 Theorem 1.3 is based on a recent work by
Bouchitt\'e, Fonseca and Mascarenhas \cite{7}. We have thus obtained the following relaxation result:

\medskip
\noindent\proclaim{Corollary 1.4} Under the hypotheses of Theorems 1.1 and
1.3(i), (ii), (iii),
  we have $\Cal F(u,\Omega)= H(u,\Omega)$ for all
$u\in BV(\Omega;\Bbb R)$.
\endproclaim
\medskip
If we require (1.8) to be satisfied locally $u$ in compact sets of
$\Bbb R$, i.e.,  for every $r>0$ there exist $C>0$ and $A\in L^1(\Omega;[0,\infty))$,
 such that
$$
 0\le f(x,u,\xi)\le C|\xi|+A(x)
$$
 for all $(x,u,\xi) \in \Omega\times\Bbb R\times\Bbb R^N$ with $|u|\le r$, then
 it can be shown that Corollary 1.4 continues to hold
for all $u\in BV(\Omega;\Bbb R)\cap L^\infty(\Omega;\Bbb R)$.
Thus, when (1.3) and (1.4) are satisfied {\it for all} $(x_0, u_0) \in \Omega\times \Bbb R$ (see also Section 8),  Corollary 1.4 improves Theorem C 
 since  conditions (1.3) and (1.4)
imply  condition (1.7), and Corollary 1.4 does not require any coercivity properties.
\smallskip
Next we extend Theorem E to integrands $f=f(x,u,\xi)$ which depend
on $x$. As we  pointed out before, there are already  several
 results in this direction, e.g. due to Ambrosio  \cite{3}
and  later extended by De Cicco  \cite{11} to BV functions, and where
 local convergence in $L^1$ is replaced by weak convergence in $BV$.
\medskip
\noindent\proclaim{Theorem 1.5}  Assume that
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ is a  Borel integrand,
$f(x,u,\cdot)$ is  convex in $\Bbb R^N$, and   for all
$x_0\in\Omega$ and $\varepsilon>0$ there exists
$\delta>0$ such that
$$|f(x_0,u,\xi)- f(x,u,\xi)|\le\varepsilon(1+f(x,u,\xi))
\tag1.9$$
for  all $x\in\Omega$
 with $|x-x_0|\le\delta$ and for all $(u,\xi)\in\Bbb R\times\Bbb R^ N$.
Suppose  also that $f(x_0,\cdot,0)$ is lower semicontinuous and
$$\limsup_{|\xi|\to0}\frac{(f(x_0,u,0)-f(x_0,u,\xi))^+}{|\xi|}
\in L^1\loc(\Bbb R;\Bbb R).$$
Then $ H(u,\Omega)\le\Cal F(u,\Omega)$ for all $u \in BV(\Omega;\Bbb R).$
\endproclaim
\medskip

 Note that  in Theorem 1.5  we do not require  $f(x,u,\cdot)$ to be positively homogeneous
 of degree one as in Theorem D of Dal Maso.
The
proof of Theorem 1.5  relies on the blow-up method of
  Fonseca and M\"uller \cite{17}, and on
the original proof
of De Giorgi, Buttazzo and  Dal Maso
 \cite{13}.

\medskip
\noindent\proclaim{Theorem 1.6} Assume that $\Omega$ is bounded and that
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ is a Borel integrand
 which satisfies $(1.8)$, with
$f(x,u,\cdot)$  convex in $\Bbb R^N$, and
$f(\cdot,u,\xi)$ continuous in $\Omega$.
Then $\Cal F(u,\Omega)\le H(u,\Omega).$

\endproclaim


\smallskip
We now turn our attention to the vectorial case, and  consider nonnegative
integrands
$$f:\Omega\times\Bbb R^d\times\Bbb R^{dN}\to [0,\infty),\qquad\text{where $d>1$.}$$
The situation is considerably more complicated, even when
$f(x,u,\cdot)$ is assumed to be convex, rather than quasiconvex, which is the
natural assumption when $d>1$ (see \cite{6}, \cite{8}, \cite{26}).
In his  book on Calculus of Variations \cite{26, Theorems 4.1.1, 4.1.2},
 Morrey extended Serrin's Theorems A and B to the vectorial case. Several years
 later, Eisen \cite{14} studied the case where $d>1$ and proved that
  Lemma 4.14 in \cite{26},
 which is the core of Theorem B, ceases to be true when $d>1$, thus placing in doubt the validity of theorem itself. In addition, he constructed
 counterexamples for Theorems A(ii) and B(iii).
 Theorem A(ii) seems to fail in the vectorial case due mainly to the
 truncation techniques   of the type
 used in Lemma 3 of \cite{30}
(see also \cite{29, pp. 30--31}) and  in our Theorem 1.1,
suitable only for the scalar case.
  On the other hand, Serrin's Theorem A(i) continues to hold in the
 vectorial case, while the validity of Theorem B(i)--(ii) when $d>1$
 remains open.
 Note that Eisen's counterexamples were both of the form
$$f=f(u,\xi)=h(u)g(\xi).$$
Thus we cannot hope to fully extend either Theorem E of
De Giorgi, Buttazzo and  Dal Maso
or our Theorems 1.1 and 1.5 to the vectorial case. However, we can prove
the following:
\medskip
\noindent\proclaim{Theorem 1.7}  Let $f$ be
a nonnegative Borel integrand. Suppose that
for all $(x_0,u_0)\in\Omega\times\Bbb R^d$ and $\varepsilon>0$
there exist
$\delta>0$ and a modulus of continuity  $\rho$,
with $\rho(s)\le C (1+ s)$ for $s>0$  and for some $C>0$, such that
$$ f(x_0,u_0,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi)) + \rho(|u-u_0|)
\tag1.10 $$
 for  all $x\in\Omega$ with $|x-x_0|\le\delta$,
 and for all $(u,\xi) \in\Bbb R^d\times\Bbb R^{dN}$.
 Assume also that either \smallskip
 \item{\rm (a)} $f(x_0,u_0,\cdot)$ is  convex in $\Bbb R^{dN}$
or \item{}
\item{\rm (b)} $f(x_0,u_0,\cdot)$ is  quasiconvex in $\Bbb R^{dN}$ and
 $$ 0\le f(x_0,u_0,\xi)\le C(|\xi|^q+1)\qquad\text{
 for all $\xi\in\Bbb R^{dN}$, }
 \tag1.11$$
where  $C>0$  and the exponent $q\ge 1$  may depend
on $(x_0,u_0)$. In addition,  if $q>1$ then assume that
$$ f(x_0,u_0,\xi)\ge \frac1C\,|\xi|^q-C\qquad\text{
 for all $\xi\in\Bbb R^{dN}$. }
 \tag1.12$$
\smallskip\noindent
Let   $u\in BV\loc(\Omega;\Bbb R^d)$, and let
$\{u_n\}$ be a sequence of functions in $ W^{1,1}\loc(\Omega;\Bbb R^d)$ which
 converges to $u$ in $L^1\loc(\Omega;\Bbb R^d)$. Then:
\smallskip
\item{\rm (i)}
$\dsize\int_\Omega f(x,u,\nabla u)\,dx
+\int_\Omega f^\infty(x,u,dC(u))
\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n,\nabla u_n)\,dx;$
\item{}
\item{\rm (ii)} if  $f=f(x,\xi)$ then
$\dsize\int_{S(u)\cap\Omega}f^\infty(x,(u^+(x)-u^-(x))\otimes \nu_u)
\,d\Cal H^{N-1}\le\liminf_{n\to\infty}
\int_\Omega f(x,\nabla u_n)\,dx.$
\endproclaim
\medskip
 Theorem 1.7 improves
Theorem A(i) of Serrin, since condition (1.10)
is significantly weaker than the corresponding (i). Moreover,
 Theorem 1.7 is also closely related to a recent
result  of Acerbi, Bouchitt\'e  and Fonseca \cite{1}
for integrands of the form $f=f(x,\xi)$, convex in  $\xi$,  where
condition (1.10) is replaced by the growth condition
$$c_1(|\xi|^p-1)\le f(x,\xi)\le C_2(|\xi|^q+1),$$
with
$$p>q\,\frac{N-1}N.$$
In the quasiconvex case we use a  result of Ambrosio and Dal Maso
 \cite{4} for functions $g=g(\xi)$ such that
$$0\le g(\xi)\le C(1+|\xi|).$$
This growth condition is of vital importance for their argument
to work. \par
When (1.11) and (1.12) hold then, by a recent result of
Kristensen \cite{22}, we can approximate $f(x_0,u_0,\xi)$
by an increasing sequence of quasiconvex functions $g_j(\xi)$ which
 grow at most linearly, and thus we can still use \cite{4}
for each $g_j$. Note that without (1.12)
$L^1\loc$ lower semicontinuity may fail even for the simplest case
 when $f=f(\xi)$. This has been shown by  Mal\'y \cite{23} for
$$f=f(\xi)=|\text{det}\, \xi|,\qquad d=N,$$
who constructed a sequence in $W^{1,N}$ which converges
to $u(x)=x$ weakly in $W^{1,p}$, where $p<N-1$, and for which lower semicontinuity
fail (see also
 Fonseca and Mal\'y \cite{16}). Thus, by Theorem 1.7 it follows
that $f(\xi)=|\text{det}\, \xi|$ {\it cannot be approximated from below
by an increasing sequence of functions $g_j(\xi)$ which
 grow at most linearly} (see also example 7.9 in \cite{22}
for a different proof).
 This is in sharp contrast with the convex case, where
it is well known that this approximation
 can always be done (see  e.g. Proposition 9.1).\par
In Theorem 1.7(ii) we have chosen to restrict ourselves to integrands
$f$ of the form $f=f(x,\xi)$ because in this case there is
 a {\it simple} integral representation formula
for the relaxation of $F$ on the jump set $S(u)$,
while,  when $f$ depends on the full set of variables and $d>1$, then the formula
is rather complicated  (see Theorem 1.10 below).

\medskip
\noindent\proclaim{Theorem 1.8} Theorem 1.7(i) still holds
if we replace condition $(1.10)$ with the following:\smallskip\noindent
for all $(x_0,u_0)\in\Omega\times\Bbb R^d$
either $f(x_0,u_0,\xi)\equiv 0$ for all $\xi\in\Bbb R^{dN}$, or
for every $\varepsilon>0$ there exist
$C_1$, $C_2$, $\delta>0$ such that
$$\align f&(x_0,u_0,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))
\tag1.13\\
f&(x,u,\xi)\ge C_1|\xi|-C_2\tag1.14\endalign$$
for  all $(x,u)\in\Omega\times\Bbb R^d$
 with $|x-x_0|+|u-u_0|\le\delta$ and for all
$\xi\in\Bbb R^{d\, N}$.
\endproclaim\medskip

 Theorem 1.8 was proven by Fonseca and M\"uller \cite{18},
under somewhat stronger hypotheses, and in the case where assumption (b) of Theorem 1.7 holds with $q=1$. The convex case can be thought of as a natural extension of Theorem A(ii) of Serrin
to the vectorial case.
\par
Theorems 1.7 and 1.8 are complemented by the following result:
\medskip
\noindent\proclaim{Theorem 1.9} Assume that the hypotheses of
Theorem 1.3  are verified in the vectorial case, with
$f(x,u,\cdot)$  quasiconvex in $\Bbb R^{dN}$. Then
Theorem 1.3(i)--(ii)  continues to hold.
Furthermore, if in Theorem 1.3(iii) we assume that  $f^\infty=f^\infty(x,\xi)$, then
$$\Cal F(u,S(u))\le \int_{S(u)\cap\Omega}f^\infty(x,(u^+(x)-u^-(x))\otimes \nu_u)
\,d\Cal H^{N-1}(x).$$
\endproclaim
\medskip
A similar extension holds for Theorem 1.6.\par
To obtain an integral representation formula for the relaxation
$\Cal F$ over the jump set $S(u)$ in the vectorial case we need a different
set of hypotheses.
\medskip
\noindent\proclaim{Theorem 1.10} Assume that $f$ is a nonnegative
 Borel integrand which satisfies $(1.8)$.
 Suppose also that
for all $x_0\in\Omega$ and $\varepsilon>0$
there exist two constants
$\delta$,  $L>0$  such that
$$ f^\infty(x_0,u,\xi)- f^\infty(x,u,\xi)
\le\varepsilon(1+f^\infty(x,u,\xi))
\tag1.15 $$
 for  all $x\in\Omega$ with $|x-x_0|\le\delta$,
 and for all $(u,\xi) \in\Bbb R^d\times\Bbb R^{dN}$; and
$$ f^\infty(x,u,\xi)- \frac{f(x,u,t\xi)}t\le
\varepsilon\left(1+\frac{f(x,u,t\xi)}t\right)
\tag1.16 $$
 for  all $x\in\Omega$, with $|x-x_0|\le\delta$,
 and for all $(u,\xi) \in\Bbb R^d\times\Bbb R^{dN}$
and $t>L$. Then
 $$\Cal F(u,S(u))\ge
\int_{S(u)\cap\Omega}h(x,u^+(x),u^-(x), \nu_u)
\,d\Cal H^{N-1}(x), \tag1.17$$
where
$$h(x_0,\lambda,\theta,\nu)
:=\inf\left\{\int_{Q_\nu}f^\infty(x_0,w(y),\nabla w(y))\,dy:\,
w\in W^{1,1}(Q_\nu),\, w|_{\partial Q_\nu(x_0,\varepsilon)}
=u_{\lambda,\theta,\nu}\right\} \tag1.18$$
and
$$u_{\lambda,\theta,\nu}(y):=
\cases\lambda &\quad\text{if $y\cdot\nu>0,$}\\
       \theta&\quad\text{if $y\cdot\nu\le0.$}\endcases   $$
Furthermore if $(1.16)$  is replaced by
$$ \left|f^\infty(x,u,\xi)- \frac{f(x,u,t\xi)}t\right|\le
\varepsilon\left(1+\frac{f(x,u,t\xi)}t\right)
\tag"$(1.16)'$" $$
and $f^\infty(\cdot,u,\xi)$ is upper semicontinuous, then
$(1.17)$ is an equality.
\endproclaim
\medskip
Here, and in what follows,
 $Q_\nu:=R_\nu(-\frac12,\frac12)^N$, where $R_\nu$ denotes a
rotation such that $R_\nu e_N=\nu$, and $Q = (-\frac12,\frac12)^N$. Also, $C$ will denote a generic constant which may vary from line to line. \par
It is not difficult to see that
 conditions $(H2)$ and $(H4)$ in Thm. 4.1.4 of \cite{7}
imply (1.16).
\medskip
In addition to the novelty of the results in this paper,
 which significantly
improve upon classical theorems in the literature,
we would like to close this section pointing out some aspects of our
  approach. One of the main tools exploited in the paper is the
{\it blow--up method}
 introduced by Fonseca and M\"uller \cite{17}, \cite{18}.
This method was first used  to deal with quasiconvex
integrands, since many
of the techniques in convex analysis available  for
the scalar case could not be easily extended to the vectorial case.
 It  turns out
 that blow--up arguments   in the scalar
 case, combined with some classical methods
 for convex integrands, may improve and simplify
some important results in the literature. Also, we use the very recent
global method of relaxation introduced by
Bouchitt\'e, Fonseca and Mascarenhas \cite{7}, to show that the
relaxed energy density may be written in terms of a Dirichlet problem.
 Most of the proofs are carried out firstly for
 $f$ which grow at most linearly in the gradient variable $\xi$.
  While this approach is standard in the convex setting, and in the 
  literature there are several
  results  which allow to approximate from below
  convex functions by
   an increasing sequence of convex  functions  which
 grow at most linearly, it was only very recently that Kristensen
  brought this idea to the vectorial setting, exploiting his approximation
   result for quasiconvex functions (see \cite{22}; see also \cite{24}).

In the presentation of the paper, and whenever it was possible,
 we have tried to treat separately the
energies corresponding to
the Lebesgue, Cantor, and Jump part  of $Du$, in order
 to better understand the corresponding scaling and the necessity and sufficiency
 of our hypotheses.
 It is interesting to observe
 that the Lebesgue and Cantor measures may be treated in a
similar fashion and, more importantly,  under the same
hypotheses on the integrand $f$.
On the other hand lower semicontinuity
 for the jump part requires  hypotheses and methods  which depart
 from the above mentioned.

 Although the hypotheses on the integrand $f$
are rather mild, they are not by any means  minimal.
 Indeed, it was not our purpose to obtain necessary conditions for lower semicontinuity, but, rather,
to find {\it simple} sufficient assumptions, which would be easy to verify
in the applications.
 It seems, however, that lower semicontinuity of $f$ in the $x$ variable
 is almost necessary, but it is not clear if it should always be uniform
in $\xi$ (at least for functionals which are allowed to vanish).
 Dal Maso's example (see \cite{9})
$$f(x,\xi)= |(\sin \omega (x),\cos \omega(x))\cdot\xi|$$
certainly seems to imply that it should.
The lower  semicontinuity of $f$ in the $u$ variable
 is not necessary, as proved by Theorem D, but
in order to drop it, stronger assumptions on the dependence on $x$
seem to be needed.



\bigskip
\noindent {\bf\S2. Proof of Theorem 1.1.}

 Throughout this work we will use often truncation arguments, and the result below will be instrumental.

\medskip
\noindent\proclaim{Proposition 2.1} {\rm (Truncation)} Let
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ be a Borel integrand
satisfying $(1.8)$.
Suppose that there exists a sequence
$(\varepsilon_k,\lambda_k,t_k,u_{0k})\in\Bbb R^4$ such that
$$\varepsilon_k\to 0^+, \qquad\lambda_k\to\lambda\in [0,\infty),
\qquad t_k\to T\in (0,\infty],\qquad
u_{0k}\to u_0\in\Bbb R,$$
and
$$\lim_{k\to \infty}\frac1{t_k}\int_{Q_\nu}
f(x_0+\varepsilon_k y,u_{0k}+\lambda_k w_k(y),t_k\nabla w_k(y))\,dy=:\ell<\infty,
$$
where $x_0\in\Omega$,
$\{w_k\}\subset W^{1,1}(Q_\nu;\Bbb R)$ converges in $L^1(Q_\nu;\Bbb R)$ to a function
$w_0\in L^\infty(Q_\nu;\Bbb R)$. \par
Let $[\lambda\,\mathop{\text{\rm essinf}}_{Q_\nu} w_0,\lambda\,
\mathop{\text{\rm  esssup}}_{Q_\nu} w_0]\subset(\alpha_1,\alpha_2)$, for
some $\alpha_1$, $\alpha_2\in\Bbb R$. Then
there exists a new sequence
$\{v_k\}\subset W^{1,1}(Q_\nu;\Bbb R)$, converging to $w_0$ in $L^1(Q_\nu;\Bbb R)$,  such
 that
$$\lim_{k\to \infty}\frac1{t_k}\int_{Q_\nu}
f(x_0+\varepsilon_k y,u_{0k}+\lambda_k v_k(y),t_k\nabla v_k(y))\,dy\le\ell
$$
and
$$u_{0k}+\lambda_k v_k(y)\in [u_0+\alpha_1,u_0+\alpha_2]\qquad\text{\rm
for $\Cal L^N$ a.e. $y\in Q_\nu$.}$$
\endproclaim
\medskip
\remark{Remark 2.2} (i) It is easy to check that the conclusion
of Proposition 2.1 still holds if we replace $Q_\nu$ by any bounded, open,
convex subset of $\Bbb R^N$ containing the origin.\par
(ii) Condition (1.8)
 can be significantly weakened if we specialize the sequences
$t_k$, $\varepsilon_k$ and $x_0$ (see Lemma 8.4).
\endremark
\medskip
\demo{Proof} Take $0<2\varepsilon< \min\{\lambda\,
\mathop{\text{\rm  essinf}}_{Q_\nu} w_0 -\alpha_1,
\alpha_2-\lambda
\,\mathop{\text{\rm  esssup}}_{Q_\nu} w_0\}$, and let $k$ be so  large that
$|u_0-u_{0k}|<\varepsilon/2$. Define
$$ \aligned
&E_k:=\{y\in Q_\nu: \,u_{0k}+\lambda_k\,w_k(y)\in[u_0 +\alpha_1,u_0+\alpha_2]\},\\
&E_k^+:=\{y\in Q_\nu: \,u_{0k}+\lambda_k\,w_k(y)>u_0+\alpha_2\},\\
&E_k^-:=\{y\in Q_\nu: \,u_{0k}+\lambda_k\,w_k(y)<u_0 +\alpha_1\},\endaligned\qquad\qquad
 v_k(y):=\cases w_k(y)\quad&\text{in $E_k$},\\
(u_0-u_{0k}+\alpha_2)/\lambda_k\quad&\text{in $E_k^+$},\\
(u_0-u_{0k} +\alpha_1)/\lambda_k\quad&\text{in $E_k^-.$}
\endcases $$
Then $v_k\in  W^{1,1}(Q_\nu;\Bbb R)$, and for $k$  large enough
$$\alpha_2>\varepsilon+|u_0-u_{0k}|
+\lambda_k \,\mathop{\text{\rm esssup}}_{Q_\nu} w_0,\qquad
\alpha_1+\varepsilon<|u_0-u_{0k}|+\lambda_k \,\mathop{\text{\rm essinf}}_{Q_\nu} w_0,\tag2.1$$
with
$$\aligned
\int_{Q_\nu}|v_k(y)-w_0(y)|dy &=
\int_{E_k}|w_k(y)-w_0(y)|dy+
\int_{E_k^+}\left(\frac{
u_0-u_{0k}+\alpha_2}{\lambda_k}-w_0(y)\right)dy
\\
&\quad +\int_{E_k^-}\left(w_0(y)-\frac{
u_0-u_{0k} +\alpha_1}{\lambda_k}\right)dy\le
||w_k-w_0||_{L^1(Q_\nu;\Bbb R)}\to 0\endaligned$$
as $k\to\infty$. Moreover
$$\aligned
&\frac1{t_k}\int_{Q_\nu}f(x_0+\varepsilon_k y,u_{0k}+\lambda_k v_k(y),t_k\nabla v_k(y))\,dy
=\frac1{t_k}\int_{E_k}
f(x_0+\varepsilon_k y,u_{0k}+\lambda_k w_k(y),t_k\nabla w_k(y))\,dy\\
&\qquad+
\frac1{t_k}\int_{E_k^+}
f(x_0+\varepsilon_k y,u_0+\alpha_2,0)\,dy+\frac1{t_k}
\int_{E_k^-}
f(x_0+\varepsilon_k y,u_0-\alpha_1,0)\,dy.\endaligned$$
By (1.8) and (2.1)
$$\aligned  0 &\le \frac1{t_k}\int_{E_k^+}
f(x_0+\varepsilon_k y,u_0+\alpha_2,0)\,dy\le
\frac C{t_k}\,\Cal L^N(E_k^+)\\
&\le \frac C{t_k}\,
\Cal L^N(\{y\in Q_\nu:\,|w_k(y)-w_0(y)|>\varepsilon/\lambda_k\})
\le \frac {C\lambda_k}{t_k\varepsilon}\,||w_k-w_0||_{L^1(Q_\nu;\Bbb R)}\to 0.
\endaligned\tag2.2$$
Similarly, we can show that the integral over $E_k^-$
approaches zero as $k\to\infty$.
\hfill\qquad\qed
\enddemo


\demo{Proof of Theorem 1.1}
Without loss of generality
we may  assume that
 $f$ is continuous, $f(x,u,\cdot)$ is convex, $f$ satisfies (1.7), (1.8), and
 for all $(x_0,u_0)\in \Omega\times \Bbb R$
 and $\varepsilon>0$ there exists
$\delta>0$ such that
$$f(x_0,u,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))
\tag2.3$$
for  all $(x,u)\in\Omega\times\Bbb R$
 with $|x-x_0|+|u-u_0|\le\delta$ and for all
$\xi\in\Bbb R^ N$. \par
Indeed, suppose that the conclusion of the theorem is true under
these additional hypotheses. By applying
Proposition 9.3 to the function $f$, and noting that (1.7) and (9.2) are equivalent if $f \geq 1$, we may find
an increasing sequence
of nonnegative
 {\it continuous} functions $f_j$ convex in $\xi$,
which satisfy (1.7), (1.8),  and (2.3),
and such that
 $f(x,u,\xi) + 1$$=\sup_{j}f_j(x,u,\xi)$.
Let $\{u_n\}\subset  W^{1,1}\loc(\Omega;\Bbb R)$ converge to
$u\in BV\loc(\Omega;\Bbb R)$ in $L^1\loc(\Omega;\Bbb R)$, and let $A\Subset\Omega$.
For any fixed $j$ 
$$\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx\ge
\liminf_{n\to\infty}
\int_A  f_j(x,u_n(x),\nabla u_n(x))\,dx - {\Cal L}^N(A) \ge H_j(u,A) - {\Cal L}^N(A),$$
where $H_j$ is the functional given in (1.2) and corresponding
 to $f_j$.
If now we let $j\to\infty$, and use Lebesgue Monotone Convergence Theorem
and Proposition 9.3, we conclude that
$$\aligned\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx &\ge
\int_A f(x,u,\nabla u)\,dx +\int_A
f^\infty(x,u,dC(u))
\\&\quad +\int_{S(u)\cap A}\left(
\int_{u^-(x)}^{u^+(x)}f^\infty(x,s,\nu_u)\,ds\right)
d\Cal H^{N-1}\\ 
& =H(u,A),\endaligned$$
where we have used the fact that $(f + 1)^{\infty} = f^{\infty}.$ The result now follows by letting $A\nearrow\Omega$, and using Lebesgue Monotone Convergence Theorem once again.

Thus, in what follows $f$ is continuous, verifies  (1.7), (1.8) and
(2.3),
 $\{u_n\}\subset  W^{1,1}\loc(\Omega;\Bbb R)$ converges to
$u\in BV\loc(\Omega;\Bbb R)$ in $L^1\loc(\Omega;\Bbb
R)$, and, without loss of generality,
we may assume that
$$\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx=
\lim_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx<\infty.\tag2.4$$
Passing to a subsequence, if necessary, we may find a
nonnegative Radon measure $\mu$ such that
$$f(x, u_n(x),\nabla u_n(x))\,\Cal L^N\lfloor\,\Omega
\buildrel\star\over\rightharpoonup\mu$$
as $n\to\infty$, weakly $\star$ in the sense of measures. \smallskip
(i) {\it (Lebesgue part)}\, We claim that
$$\frac{d\mu}{d\Cal L^N}(x_0)=\lim_{\varepsilon\to 0^+}
\frac{\mu(Q(x_0,\varepsilon))}{\varepsilon^N}\ge
f(x_0, u_n(x_0),\nabla u_n(x_0))\qquad\text{for
$\Cal L^N$ a.e. $x_0\in \Omega$},\tag2.5$$
where $Q(x_0,\varepsilon):=x_0+\varepsilon\,Q$.
If (2.5) holds, then the conclusion of the theorem follows immediately.
Indeed, let $\varphi\in C_0(\Omega;\Bbb R)$, $0\le\varphi\le1$. We have
$$\aligned
\lim_{n\to\infty}
\int_\Omega f(x,u_n,\nabla u_n)\,dx&\ge
\liminf_{n\to\infty}
\int_\Omega\varphi\, f(x,u_n,\nabla u_n)\,dx=
\int_\Omega\varphi\,d\mu\\
&\ge\int_\Omega \varphi\,
\frac{d\mu}{d\Cal L^N}\,dx\ge
\int_\Omega\varphi\, f(x,u,\nabla u)\,dx.
\endaligned$$
By letting $\varphi\to 1$, and using
Lebesgue Dominated
 Convergence Theorem,
 we obtain the desired result.    Thus, to conclude the proof of the theorem
  in what concerns the absolutely continuous part, it suffices to show
 (2.5).\par
Fix $x_0\in\Omega$
 such that
$$\frac{d\mu}{d\Cal L^N}(x_0)=\lim_{\varepsilon\to 0^+}
\frac{\mu(Q(x_0,\varepsilon))}{\varepsilon^N}<\infty,\qquad
\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon^{N+1}}\int_{Q(x_0,\varepsilon)}
|u(x)-u(x_0)-\nabla u(x_0)(x-x_0)|dx=0.
\tag2.6$$
Choosing $\varepsilon_k\to 0^+$ such that $\mu(\partial Q(x_0,\varepsilon_k))=0$, then
$$\aligned
\lim_{k\to \infty}
\frac{\mu(Q(x_0,\varepsilon_k))}{\varepsilon^N_k}&=
\lim_{k\to \infty}\lim_{n\to \infty}\frac{1}{\varepsilon^N}\int_{Q(x_0,\varepsilon_k)}
f(x,u_n,\nabla u_n)\,dx\\
&=\lim_{k\to \infty}\lim_{n\to \infty}\int_{Q}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kw_{n,k}(y),\nabla w_{n,k}(y))\,dy,
\endaligned
$$
where
$$w_{n,k}(y):=\frac{u_n(x_0+\varepsilon_ky)-u(x_0)}{\varepsilon_k}.$$
Clearly $w_{n,k}\in  W^{1,1}(Q;\Bbb R)$, and  by (2.6),
$\lim\limits_{k\to \infty}\lim\limits_{n\to \infty}||w_{n,k}-w_0||_{L^1(Q;\Bbb R)}=0$,
where $w_0(y):=\nabla u(x_0)y$.
By a standard diagonalization argument, we may extract a subsequence
$w_k:=w_{n_k,k}$ which converges to $w_0$ in $L^1(Q;\Bbb R)$ and such that
$$
\frac{d\mu}{d\Cal L^N}(x_0)=
\lim_{k\to \infty}\int_{Q}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kw_k(y),\nabla w_k(y))\,dy.\tag2.7
$$
Fix $\varepsilon>0$ and let $\delta$ be provided by (1.7).
By Proposition 2.1, with
$$\lambda_k:=\varepsilon_k,\qquad t_k:\equiv 1,\qquad u_{0k}:\equiv u(x_0),
\qquad -\alpha_1=\alpha_2:=\delta,$$
we may find a new sequence
$\{v_k\}\subset W^{1,1}(Q;\Bbb R)$, still convergent to $w_0$ in $L^1(Q;\Bbb R)$,  such
 that
$$\frac{d\mu}{d\Cal L^N}(x_0)\ge
\liminf_{k\to\infty}
\int_{Q}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kv_k(y),\nabla v_k(y))\,dy,$$
and $|\varepsilon_k v_k(y)|\le \delta$ for $\Cal L^N$ a.e. $y\in Q$.
Since $\varepsilon_k\to 0$, by (1.7) we obtain
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon\ge\liminf_{k\to\infty}
\int_{Q}
f(x_0,u(x_0),\nabla v_k(y))\,dy.$$
We can now apply Serrin's Theorem A(i) to the integrand
$g(\xi):= f(x_0,u(x_0),\xi)$ to conclude that

$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon\ge
f(x_0,u(x_0),\nabla u(x_0)).$$
The result follows by letting $\varepsilon \to 0^+$.
\hfill\qquad\qed
\medskip
(ii) {\it (Cantor part)}\,  The proof for the Cantor part is somewhat similar
 to the previous one, and we will only indicate the main differences.
 The inequality (2.5) is now replaced by
$$\frac{d\mu}{d|C(u)|}(x_0)\ge
f^\infty\left(x_0, u(x_0),\frac{d C(u)}{d|C(u)|}(x_0)\right)\qquad\text{for
$C(u)$ a.e. $x_0\in \Omega$},$$
where (see \cite{2})
$$\frac{d C(u)}{d|C(u)|}(x_0)=a_u(x_0)\nu_u(x_0),\tag2.8$$
with $a_u(x_0)\in\Bbb R$ and $\nu_u(x_0)\in S^{N-1}:=\{x\in\Bbb R^N:\,|x|=1\}$.
For simplicity of the notation, from now on we will write $a$ and $\nu$ to designate $a_u(x_0)$ and
$\nu_u(x_0)$, respectively.
It is known  (see  \cite{4}, \cite{7}, \cite{18}) that
for $C(u)$ a.e. $x_0\in \Omega$ the following hold
$$\aligned
\frac{d\mu}{d|C(u)|}(x_0)&=\lim_{\varepsilon\to 0^+}
\frac{\mu(Q_\nu(x_0,\varepsilon))}{|Du|(Q_\nu(x_0,\varepsilon))}<\infty,\qquad
\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon^N}\int_{Q_\nu(x_0,\varepsilon)}|u(x)-u(x_0)|dx=0,\\
&\lim_{\varepsilon\to 0^+}\frac{|Du|(Q_\nu(x_0,\varepsilon))}{\varepsilon^{N-1}}=0,\qquad
\lim_{\varepsilon\to 0^+}\frac{|Du|(Q_\nu(x_0,\varepsilon))}{\varepsilon^N}=\infty.
\endaligned\tag2.9$$
Fix $x_0\in\Omega$ so that (2.9) holds. By Lemma 3.9 in \cite{7}
(see also Theorem  2.3 in \cite{4}) there exist
 $\varepsilon_k\to 0^+$ such that $\mu(\partial Q_\nu(x_0,\varepsilon_k))=0$, and a non decreasing
 function $\Psi:(-\frac12,\frac12)\to\Bbb R$ such that the followings
 hold:
$$\aligned \Psi(\tfrac12-0)&-\Psi(-\tfrac12+0)=1,\qquad
\int_{-\frac12}^{\frac12}\Psi(s)\,ds=0,\\
z_k(y):&=\frac{u(x_0+\varepsilon_ky)-\frac{1}{\varepsilon_k^N}\frac{1}{\varepsilon_k^N}\int_{Q_\nu}u(x_0+\varepsilon_kz)dz}{\lambda_k}\to
w_0(y):=\Psi(y\cdot\nu)\,a\quad\text{in $L^1(Q_\nu;\Bbb R)$,}\\
&\lim_{k\to\infty}|D z_k|(Q_\nu)=|D w_0|(Q_\nu),
\endaligned\tag2.10$$
where, by (2.9), $\lambda_k:=|Du|(Q_\nu(x_0,\varepsilon_k))/\varepsilon_k^{N-1}\to 0$ and
$t_k:=\lambda_k/\varepsilon_k\to \infty$ as $k\to\infty$. Then
$$\aligned
\frac{d\mu}{d|C(u)|}(x_0)=\lim_{k\to \infty}
\frac{\mu(Q_\nu(x_0,\varepsilon_k))}{|Du|(Q_\nu(x_0,\varepsilon_k))}&=
\lim_{k\to \infty}\lim_{n\to \infty}\frac1{|Du|(Q_\nu(x_0,\varepsilon_k))}
\int_{Q_\nu(x_0,\varepsilon_k)}
f(x,u_n,\nabla u_n)\,dx\\
&=\lim_{k\to \infty}\lim_{n\to \infty}\frac1{t_k}\int_{Q_\nu}
f(x_0+\varepsilon_k y,u_{0n,k}+\lambda_k w_{n,k}(y),t_k\nabla w_{n,k}(y))\,dy,
\endaligned
$$
where
$$w_{n,k}(y):=\frac{u_n(x_0+\varepsilon_ky)-u_{0n,k}}{\lambda_k},\qquad
u_{0n,k}:=\frac{1}{\varepsilon_k^N}\int_{Q_\nu}u_n(x_0+\varepsilon_kz)dz.$$
Clearly $w_{n,k}\in  W^{1,1}(Q_\nu;\Bbb R)$, and  by (2.10),
(2.9)$_2$, and
the fact that $\{u_n\}$ converges to $u$
 in $L^1$,
$$\lim\limits_{k\to \infty}\lim\limits_{n\to \infty}||w_{n,k}-w_0||_{L^1(Q_\nu;\Bbb R)}=0,\qquad
\lim\limits_{k\to \infty}\lim\limits_{n\to \infty}u_{0n,k}=u(x_0).$$
By a standard diagonalization argument, we may extract two subsequences
$\{w_k:=w_{n_k,k}\} $, $\{u_{0k}:=u_{0n_k,k}\}$, which converge
 to $w_0$ in $L^1(Q_\nu;\Bbb R)$ and  $u(x_0)$, respectively,
such that
$$
\frac{d\mu}{d|C(u)|}(x_0)=
\lim_{k\to \infty}\frac1{t_k}\int_{Q_\nu}
f(x_0+\varepsilon_k y,u_{0k}+\lambda_k w_k(y),t_k\nabla w_k(y))\,dy.\tag2.11
$$
We can now continue as in part (i), using Proposition 2.1 and then (1.7)
 to conclude that
$$(1+\varepsilon)\frac{d\mu}{d|C(u)|}(x_0)+\varepsilon
\ge\liminf_{k\to\infty}
\frac1{t_k}\int_{Q_\nu}
f(x_0,u(x_0),t_k\nabla v_k(y))\,dy.\tag2.12$$
Due to the presence of
the sequence $t_k$,  we cannot  apply directly Serrin's Theorem A(i) to
the integrand
$g(\xi):= f(x_0,u(x_0),\xi)$ as we did in part(i). Although
  the adaptations to the present setting are quite straightforward,
 here we present an alternative proof which can be extended
 to the vectorial case and to quasiconvex functions.
Assume, for simplicity, that $\nu=e_N$ and
construct a sequence of
smooth functions $v_h(y)=\bar v_h(y_N)$ such that
$$||v_h-w_0||_{L^1(Q;\Bbb R)}\le\frac1h\qquad \text{and}\qquad
|\nabla v_h|(Q)-|Dw_0|(Q)\to 0.$$
as $h\to\infty$. Since $v_h$ depends only on $y_N$, its trace on
$\partial Q$ agrees with the trace of a function
$$A_h\,y+p(y),\qquad A_h:=(\bar v_h(1/2)-\bar v_h(-1/2))\otimes
e_N=\nabla v_h(Q),$$
where $p$ is $Q$--periodic. Choose open sets $\Omega_i$, $i=1,2,3$, such that $\Omega_1\Subset
\Omega_2\Subset \Omega_3\Subset Q$. Since $g$ satisfies (1.8), by
Lemma 2.5 of Ambrosio and Dal Maso \cite{4}
we may find a new sequence
$$v_{h,k}(y)=\varphi(y) w_k(y)+(1-\varphi(y))v_h(y) $$
such that
$$
\frac1{t_k}\int_{\Omega_3} g(t_k\nabla w_k)\,dy+
\frac1{t_k}\int_{Q\backslash\Omega_1} g(t_k\nabla v_h)\,dy
+\frac{4C}{\delta}\int_Q|w_k-v_h|\,dy+\frac1k
\ge \frac1{t_k}\int_Q g(t_k\nabla v_{h,k})\,dy$$
where $\varphi$ is a cut--off function such that
$\varphi\equiv 1$ in a neighborhood of $\Omega_2$,
$\varphi\equiv 0$ in a neighborhood of $\Bbb R^N \backslash \Omega_3$, and
 $\delta<$ dist $ (\Omega_2,\partial \Omega_3)$.
By virtue of the quasiconvexity of $g$, together with the growth
 (1.8)
(recall that in the scalar case quasiconvexity is equivalent to convexity),
 we obtain
$$
\frac1{t_k}\int_Q g(t_k\nabla w_k)\,dy+
C\left(\frac{\Cal L^N(Q\backslash\Omega_1)}{t_k}+|\nabla v_h|(Q\backslash\Omega_1)\right)
+\frac{4C}{\delta}\int_Q|w_k-v_h|\,dy+\frac1k\ge
\frac1{t_k} g(t_k\nabla v_h(Q)).$$
Letting $k\to\infty$ gives
$$
\lim_{k\to\infty}\frac1{t_k}\int_{Q} g(t_k\nabla w_k)\,dy+
C|\nabla v_h|(Q\backslash\Omega_1)
+\frac{4C}{\delta}\int_Q|w_0-v_h|\,dy\ge  g^\infty(\nabla v_h(Q));$$
hence, taking the limit as  $h\to\infty $, we conclude that
$$\lim_{k\to\infty}\frac1{t_k}\int_{Q} g(t_k\nabla w_k)\,dy+
C|D w_0|(Q\backslash\Omega_1)\ge g^\infty(\nabla w_0(Q)),$$
where we used the continuity of $g^\infty$ (see Proposition 9.1), and
the fact that $|\nabla v_n|(Q)\to|Dw_0|(Q)$.
If we now let $\Omega_1\nearrow Q$ we get
$$\lim_{k\to\infty}\frac1{t_k}\int_{Q}
 g(t_k\nabla w_k)\,dy\ge g^\infty(a\, \nu);$$
thus,
$$(1+\varepsilon)\frac{d\mu}{d|C(u)|}(x_0)+\varepsilon\ge
f^\infty\left(x_0,u(x_0),\frac{d C(u)}{d|C(u)|}(x_0)\right).$$
It  suffices now to let $\varepsilon \to 0^+$.
\hfill\qquad\qed
\medskip
(iii){\it (Jump part)} To complete the proof of Theorem 1.1 it remains to show that
$$\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\ge
\int_{u^-(x_0)}^{u^+(x_0)}f^\infty(x_0, s,\nu)\,ds\qquad\text{for
$\Cal H^{N-1}$ a.e. $x_0\in S(u)$},$$
where  $\nu=\nu_u(x_0)$ is the normal to $S(u)$.
It is known that (see  \cite{7})
for
$\Cal H^{N-1}$ a.e. $x_0\in S(u)$
$$\aligned
\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)&=\lim_{\varepsilon\to 0^+}
\frac{\mu(Q_\nu(x_0,\varepsilon))}{\varepsilon^{N-1}}<\infty,\\
\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon^N}\int_{Q_\nu^+(x_0,\varepsilon)} |u(x)-u^+(x_0)|dx&=0,\qquad
\lim_{\varepsilon\to 0^+}\frac{1}{\varepsilon^N}\int_{Q_\nu^-(x_0,\varepsilon)}|u(x)-u^-(x_0)|dx=0,
\endaligned\tag2.13$$
where $Q_\nu^+(x_0,\varepsilon):=\{y\in Q_\nu(x_0,\varepsilon):
\,y\cdot\nu>0\}$ and $Q_\nu^-(x_0,\varepsilon):=\{y\in Q_\nu(x_0,\varepsilon):
\,y\cdot\nu<0\}$.
Fix $x_0\in S(u)$ such that
(2.13) holds, and choose
a sequence
 $\varepsilon_k\to 0^+$ with $\mu(\partial Q_\nu(x_0,\varepsilon_k))=0$. Then
$$\aligned
\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)&=\lim_{k\to \infty}
\frac{\mu(Q_\nu(x_0,\varepsilon_k))}{\varepsilon_k^{N-1}}=
\lim_{k\to \infty}\lim_{n\to \infty}\frac1{\varepsilon_k^{N-1}}
\int_{Q_\nu(x_0,\varepsilon_k)}
f(x,u_n,\nabla u_n)\,dx\\
&=\lim_{k\to \infty}\lim_{n\to \infty}
\int_{Q_\nu}\varepsilon_k f\left(x_0+\varepsilon_k y, w_{n,k}(y),
\frac1{\varepsilon_k}\nabla w_{n,k}(y)\right)\,dy,
\endaligned
$$
where $w_{n,k}(y):=u_n(x_0+\varepsilon_ky)$.
Clearly $w_{n,k}\in  W^{1,1}(Q_\nu;\Bbb R)$, and  by (2.13) together
with
the fact that $u_n$ converges to $u$
 in $L^1$,
$$\lim\limits_{k\to \infty}\lim\limits_{n\to \infty}||w_{n,k}-w_0||_{L^1(Q_\nu;\Bbb R)}=0,\qquad
\text{where}\qquad w_0(y):=\cases u^+(x_0)&\quad\text{if
$y\cdot\nu>0$},\\
                                  u^-(x_0)&\quad\text{if $y\cdot\nu\le0$}.
\endcases$$
As before, by a standard diagonalization argument we may extract a subsequence
$\{w_k:=w_{n_k,k}\}$  converging to $w_0$ in $L^1(Q_\nu;\Bbb R)$, with
$$
\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)=\lim_{k\to \infty}\int_{Q_\nu}
\varepsilon_k f\left(x_0+\varepsilon_k y, w_k(y),\frac1{\varepsilon_k}
\nabla w_k(y)\right)\,dy.
\tag2.14$$
Fix  $\varepsilon>0$. By (2.3)  for each $u_1\in [u^-(x_0),u^+(x_0)]$ there is
$\delta_{u_1}>0$ such that
$$f(x_0,u,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))$$
for  all  $|x-x_0|\le \delta_{u_1}$,   $|u-u_1|\le\delta_{u_1}$, and for all
$\xi\in\Bbb R^ N$.
Since
$$\bigcup_{u_1\in[u^-(x_0),u^+(x_0)]} B(u_1,\delta_{u_1})
\supset [u^-(x_0),u^+(x_0)],$$
we may find a finite subcovering
$$\bigcup_{i=1}^M B(u_i,\delta_i)\supset [u^-(x_0),u^+(x_0)].$$
Set $\delta:=\min\{\delta_1,\ldots,\delta_M,\delta^+,\delta^-\}$,
where $\delta^\pm$ are  provided by  (2.3) corresponding
 to the points $(x_0,u^\pm(x_0))$, respectively. Then
$$f(x_0,u,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))\tag2.15$$
for  all  $|x-x_0|\le \delta$, $u\in
[u^-(x_0)-\delta,u^+(x_0)+\delta] $, and for all
$\xi\in\Bbb R^ N$.
\par
By Proposition 2.1, with
$$\lambda_k:\equiv 1,\qquad t_k:=1/\varepsilon_k,\qquad u_{0k}:\equiv 0,\qquad
\alpha_1=u^-(x_0)-\delta,\qquad \alpha_2:=u^+(x_0)+\delta,$$
there exists a new sequence
$v_k\in W^{1,1}(Q_\nu;\Bbb R)$,  convergent to $w_0$ in $L^1(Q_\nu;\Bbb R)$,  such
 that $v_k(y)\in [u^-(x_0)-\alpha_1,u^+(x_0)+\alpha_2]$, and by (2.14) and
 (2.15) we have
$$(1+\varepsilon)\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
\ge\liminf_{k\to\infty}
\int_{Q_\nu}\varepsilon_k
f\left(x_0,v_k(y),\frac1{\varepsilon_k}\nabla v_k(y)\right)\,dy.$$
Since $h(u,\xi):=f(x_0,u,\xi)$  is continuous,
by  an approximation result due to   Ambrosio \cite{3} we can write
$$h(u,\xi)=\sup_{i\in\Bbb N}[a_i(u)+b_i(u)\cdot\xi]^+$$
where the functions $a_i:\Bbb R\to \Bbb R$ and $b_i:\Bbb R\to \Bbb R^N$
are bounded and continuous.
It is not difficult to see that
$$h^\infty(u,\xi)=\sup_i[b_i(u)\cdot\xi]^+.$$
Therefore (see [12, Lemma 6])
$$
\int_{u^-(x_0)}^{u^+(x_0)}\!\!h^\infty(s,\nu)\,ds = 
\sup_{j\in \Bbb N}\sup\biggl\{\sum_{i=1}^j
\int_{u^-(x_0)}^{u^+(x_0)} \!\!\!\psi_i(s)[b_i(s)\cdot\nu]^+\,ds:
 \psi_i \in C_0^{\infty}((u^-(x_0)^,u^+(x_0));[0,1]), \sum_{i=1}^j \psi\leq 1 \biggr\}.
\tag2.16
$$
Fix $j \in \Bbb N$, $\psi_1, \ldots, \psi_j, $ as in (2.16), and let $\varphi \in C_0^{\infty}(Q_\nu; [0,1])$. 
We have
$$\aligned
(1+\varepsilon)\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
&\ge\liminf_{k\to\infty}\sum_{i=1}^j
\int_{Q_\nu} \varphi(y) \varepsilon_k\psi_i(v_k(y))\left[a_i(v_k(y))+\frac1{\varepsilon_k}\,b_i(v_k(y))\cdot\nabla v_k(y)\right]^+\,dy\\
& \ge \liminf_{k\to\infty}\sum_{i=1}^j
\int_{Q_\nu} \varphi(y) \psi_i(v_k(y))\,\left[b_i(v_k(y))\cdot\nabla v_k(y)\right]^+\,dy,
\endaligned
$$
where we have used the inequality  $(\alpha+\beta)^+\ge(\beta)^+-|\alpha|$ for $\alpha$, $\beta\in\Bbb R$, Lebesgue
Dominated Convergence Theorem,   and the fact that
$$\varepsilon_k |a_i(v_k(y))|\le ||a_i||_{L^\infty(\Bbb R)}
\,\varepsilon_k \to 0.
$$
By a result of De Cicco \cite{12, Theorem 1}, we have
$$
(1+\varepsilon)\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
\ge\sum_{i=1}^j \int_{S(w_0)} \varphi(y)
\int_{u^-(x_0)}^{u^+(x_0)}\psi_i(s)\,[b_i(s)\cdot\nu]^+\,ds\,d\Cal H^{N-1}(y),$$
and taking $\varphi \nearrow 1$, we obtain
$$
\aligned
(1+\varepsilon)\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
&\ge\sum_{i=1}^j \int_{S(w_0)}
\int_{u^-(x_0)}^{u^+(x_0)}\psi_i(s)\,[b_i(s)\cdot\nu]^+\,ds\,d\Cal H^{N-1}(y)\\
&= \sum_{i=1}^j 
\int_{u^-(x_0)}^{u^+(x_0)}\psi_i(s)\,[b_i(s)\cdot\nu]^+\,ds.
\endaligned
$$
In view of (2.16), the proof is concluded by  taking the supremum over all $j\in \Bbb N$, $\psi_1, \ldots, \psi_j$, and  letting $\varepsilon\to 0^+$.
\hfill\qquad\qed
\enddemo

\bigskip

\noindent {\bf\S3. Proof of Theorem 1.3.}
\medskip
(i){\it  (Lebesgue part)} Assume first that $f$
 is Carath\'eodory.
Consider $\Cal F:BV(\Omega;\Bbb R)\times\Cal A(\Omega)\to[0,\infty]$,
where $\Cal A(\Omega)$ stands for the family of open subsets of $\Omega$. It can be proved
that $\Cal F(u;\cdot)$ is the restriction to $\Cal A(\Omega)$ of a Radon measure,
$\Cal F(\cdot;A)$ is $L^1(A)$ lower semicontinuous, and
$$0\le \Cal F(u,A)\le C\left(\Cal L^N(A)+|Du|(A)\right).$$
For a proof we refer to  Lemma 4.1.2 of \cite{7}
(see also \cite{5}, \cite{18}).
Let $\Cal F_1(u,A):=\Cal F(u,A)+|Du|(A)$. By Theorem 3.7 of  \cite{7}
we have
$$\frac{d\Cal F_1(u,\cdot)}{d\Cal L^N}(x_0)=\frac{d\Cal F(u,\cdot)}{d\Cal L^N}(x_0)
+|\nabla u(x_0)|=f_1(x_0,u(x_0),\nabla u(x_0))\qquad\text{for
$\Cal L^N$ a.e. $x_0\in \Omega$},$$
where
$$f_1(x_0,u_0,\xi):=\limsup_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\,\inf\left\{\Cal F_1(v,Q(x_0,\varepsilon)):\,
v\in BV(Q(x_0,\varepsilon)),\, v|_{\partial Q(x_0,\varepsilon)}=u_0+\xi\cdot (x-x_0)\right\}.$$
Thus, the proof of part (i) is completed provided  we show that
$$f_1(x_0,u_0,\xi)\le f(x_0,u_0,\xi)+|\xi|$$
for
$\Cal L^N$ a.e. $x_0\in \Omega$ and for all $(u_0,\xi)\in\Bbb R\times\Bbb R^N$. Clearly
$$f_1(x_0,u_0,\xi)\le\limsup_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\,\Cal F(u_0+\xi\cdot (x-x_0),Q(x_0,\varepsilon))+|\xi|\le
\limsup_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\int_{Q(x_0,\varepsilon)} f(x,u_0+\xi\cdot (x-x_0),\xi)\,dx
+|\xi|.$$
Since $f$ is Carath\'eodory, by the Scorza-Dragoni Theorem for each $i\in\Bbb N$ there exists a compact set
$K_i\subset\Omega$, with $\Cal L^N(\Omega\backslash K_i)\le1/i$, such that
$f:K_i\times\Bbb R\times\Bbb R^N\to[0,\infty]$ is continuous. Let $K_i^*$ be the
set of Lebesgue points of $\chi_{K_i}$, and set $\omega:=\cup_{i}^\infty (K_i\cap K_i^*)$.
 Then
$$\Cal L^N(\Omega\backslash\omega)\le \Cal L^N(\Omega\backslash K_i)\le\frac1i
\to 0\quad\text{as $i\to\infty$}.$$
If $x_0\in \omega$ then  $x_0\in K_i\cap K_i^*$ for some index $i$.
Since $g(x):=f(x,u_0+\xi(x-x_0),\xi)$ is continuous over $ K_i$, given $\delta>0$
there exists $\eta>0$ such that
$g(x)\le g(x_0)+\delta$ for all $x\in K_i$ with $|x-x_0|\le\eta$.
Therefore, by (1.8) we have
$$\aligned f_1(x_0,u_0,\xi)-|\xi|&\le (f(x_0,u_0,\xi)+\delta)\,\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q(x_0,\varepsilon)\cap K_i)}{\varepsilon^N}+C(1+|\xi|)
\,\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q(x_0,\varepsilon)\backslash K_i)}{\varepsilon^N}\\
&=f(x_0,u_0,\xi)+\delta,\endaligned$$
where we have used the fact that $x_0$ is a  Lebesgue point of $\chi_{K_i}$.
By letting $\delta\to 0^+$ we obtain the desired inequality.
The argument for the case where $f(\cdot,\cdot,\xi)$ is upper
semicontinuous is very similar to the
one used in Theorem 1.3(ii) below, and therefore we omit the details.
\hfill\qquad\qed

\medskip
(ii){\it (Cantor part)} By Lemma 3.9 of
\cite{7},
for
$C(u)$ a.e. $x_0\in \Omega$ there exist a double indexed sequence $\{t_\varepsilon^{(k)},
u_\varepsilon^{(k)}\}$ such that for every $k$
$$t_\varepsilon^{(k)}\to\infty,\qquad \varepsilon\,t_\varepsilon^{(k)}\to 0^+,
\qquad u_\varepsilon^{(k)}\to u(x_0)\quad\text{as $\varepsilon\to
0^+$,}\tag3.1$$
and
$$\aligned &\frac{d\Cal F_1(u,\cdot)}{d|C(u)|}(x_0)=
\frac{d\Cal F(u,\cdot)}{d|C(u)|}(x_0)+|a|
\\
&=\lim_{k\to \infty}\limsup_{\varepsilon\to 0^+}
\frac{\inf\left\{\Cal F_1(v,Q_\nu^{(k)}(x_0,\varepsilon)):\,
v\in BV(Q_\nu^{(k)}(x_0,\varepsilon)),\, v|_{\partial Q_\nu^{(k)}(x_0,\varepsilon)}
=u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot (x-x_0)\right\}}{k^{N-1}\varepsilon^Nt_\varepsilon^{(k)}},
\endaligned$$
where
$\dfrac{d C(u)}{d|C(u)|}(x_0)=a\nu$,
 $a=a(u,x_0)\in\Bbb R$, $\nu=\nu(u,x_0)\in S^{N-1}$,
 $Q^{(k)}_\nu(x_0,\varepsilon):=x_0+\varepsilon\,Q^{(k)}_\nu$
with
$$Q^{(k)}_\nu:=R_\nu\left(\left(-\frac k2,\frac k2\right)^{N-1}\times
\left(-\frac12,\frac12\right)\right),$$
where $R_\nu$ denotes a rotation such that
$R_\nu e_N=\nu$.
Take $x_0\in\Omega$,  so that
all the limits above exist and are finite. Then
$$\aligned \frac{d\Cal F(u,\cdot)}{d|C(u)|}(x_0)+|a|&\le
\lim_{k\to \infty}\limsup_{\varepsilon\to 0^+}
\frac1{k^{N-1}\varepsilon^Nt_\varepsilon^{(k)}}
\,\Cal F_1(u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot(x-x_0),Q^{(k)}_\nu(x_0,\varepsilon))\\
&\le\lim_{k\to \infty}\limsup_{\varepsilon\to 0^+}
\frac1{k^{N-1}\varepsilon^Nt_\varepsilon^{(k)}}
\int_{Q^{(k)}_\nu(x_0,\varepsilon)}
f(x,u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot (x-x_0),
t_\varepsilon^{(k)}a\nu)\,dx+|a|.
\endaligned\tag3.2$$
By Proposition 9.1, (9.1), and (1.8),
$$ \aligned \frac{f(x,u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu(x-x_0)
,t_\varepsilon^{(k)}a\nu)}
{t_\varepsilon^{(k)}}&\le f^\infty (x,u_\varepsilon^{(k)}
+t_\varepsilon^{(k)}a\nu\cdot(x-x_0),a\nu)+
\frac{f(x,u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot(x-x_0),0)}
{t_\varepsilon^{(k)}}\\
&\le f^\infty (x,u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot(x-x_0),a\nu)+
\frac{C}{t_\varepsilon^{(k)}}.\endaligned$$
Therefore, by (3.1) and (3.2),
$$\aligned\frac{d\Cal F(u,\cdot)}{d|C(u)|}(x_0)&\le
\lim_{k\to \infty}\limsup_{\varepsilon\to 0^+}
\frac1{k^{N-1}\varepsilon^N}
\int_{Q^{(k)}_\nu(x_0,\varepsilon)}
f^\infty (x,u_\varepsilon^{(k)}+t_\varepsilon^{(k)}a\nu\cdot(x-x_0),a\nu)dx
\\
&=\lim_{k\to \infty}\limsup_{\varepsilon\to 0^+}\frac1{k^{N-1}}
\int_{Q_\nu^{(k)}}
f^\infty(x_0+\varepsilon y,u_\varepsilon^{(k)}+\varepsilon\,
t_\varepsilon^{(k)}a\nu\cdot y,a\nu)\,dy.
\endaligned$$
Since  the function
$f^\infty(\cdot,\cdot,a\nu)$ is upper semicontinuous,
 given $\delta>0$
there exists $\eta>0$ such that
$f^\infty(x,u,a\nu)$  $\le f^\infty(x_0,u(x_0),a\nu)+\delta$
for all
 $|x-x_0|\le\eta$ and $|u-u(x_0)|\le\eta$.
By (3.1), for each fixed $k$ if $\varepsilon$ is small enough
then $x_0+\varepsilon y\in B(x_0,\eta)$ and
$u_\varepsilon^{(k)}+\varepsilon\,t_\varepsilon^{(k)}a\nu\cdot y\in [u(x_0)-\eta,
u(x_0)+\eta]$ for all $y\in Q_\nu^{(k)}$. Hence
$$
\int_{Q_\nu^{(k)}}
f^\infty(x_0+\varepsilon y,u_\varepsilon^{(k)}+\varepsilon\,
t_\varepsilon^{(k)}a\nu\cdot y,a\nu)\,dy\le
(f^\infty(x_0,u(x_0),a\nu)+\delta)\,k^{N-1}$$
and, in turn,
$$\frac{d\Cal F(u,\cdot)}{d|C(u)|}(x_0)\le f^\infty(x_0,u(x_0),a\nu)+\delta.$$
We now let $\delta\to 0^+$.
\hfill\qquad\qed

\medskip
(iii) {\it (Jump part)}
By Theorem 3.7 of  \cite{7},
for
$\Cal H^{N-1}$ a.e. $x_0\in S(u)$
$$\aligned \frac{d\Cal F_1(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)&=
\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)+|u^+(x_0)-u^-(x_0)|
\\
&=\limsup_{\varepsilon\to 0^+}
\frac{\inf\left\{\Cal F_1(v,Q_\nu(x_0,\varepsilon)):\,
v\in BV(Q_\nu(x_0,\varepsilon)),\, v|_{\partial Q_\nu(x_0,\varepsilon)}
=w_0\right\}}{\varepsilon^{N-1}},
\endaligned$$
where $\nu=\nu_u(x_0)$ is the normal to $S(u)$ and
$$w_0(y):=\cases u^+(x_0)&\quad\text{if $y\cdot\nu>0,$}\\
                                  u^-(x_0)&\quad\text{if $y\cdot\nu\le0$}.\endcases   $$
Take $x_0\in S(u)$ so that
all the limit above exists and is finite.
Then
$$\aligned \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
+|u^+(x_0)-u^-(x_0)|\le
\limsup_{\varepsilon\to 0^+}
\frac{\Cal F(w_0,Q_\nu(x_0,\varepsilon))}{\varepsilon^{N-1}}
\,+|u^+(x_0)-u^-(x_0)|\endaligned.$$
In what follows we  assume  for simplicity that $x_0=0$ and $\nu=e_N$,  and we set
$$u_{n,\varepsilon}(x_N):=
\cases  u^+(x_0)\quad&\text{if $x_N\ge \varepsilon/2n$}\\
(u^+(x_0)-u^-(x_0))\dfrac{n}{\varepsilon}x_N+
\dfrac{u^+(x_0)+u^-(x_0)}2\quad&\text{if $-\varepsilon/2n\le x_N\le \varepsilon/2n$}\\
u^-(x_0)\quad&\text{if $x_N\le -\varepsilon/2n.$}
\endcases$$
Clearly $||u_{n,\varepsilon}-w_0||_{L^1(Q_\nu(x_0,\varepsilon))}\to 0$
as $n\to\infty$; thus,
by a standard diagonalization argument
$$\aligned\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
&\le\lim_{\varepsilon\to 0^+}
\liminf_{n\to \infty}\frac1{\varepsilon^{N-1}}\int_{Q_\nu(x_0,\varepsilon)}
 f(x, u_{n,\varepsilon}(x_N),0,\ldots,0,u'_{n,\varepsilon}(x_N))\,dx
\\&\le\liminf_{k\to \infty}\int_{Q_\nu}
\varepsilon_k f\left(x_0+\varepsilon_k y, v_k(y_N),0,\ldots,0,\frac1{\varepsilon_k}
v'_k(y_N)\right)\,dy,
\endaligned$$
where
$$v_k(y_N)=
\cases  u^+(x_0)\quad&\text{if $y_N\ge 1/2n_k$}\\
(u^+(x_0)-u^-(x_0))n_k \,y_N+
\dfrac{u^+(x_0)+u^-(x_0)}2\quad&\text{if $-1/2n_k\le y_N\le 1/2n_k$}\\
u^-(x_0)\quad&\text{if $y_N\le -1/2n_k$}
\endcases$$
and $n_k\to\infty$ as $k\to\infty$.
By Proposition 9.1, (9.1), and (1.8), we have
$$ \aligned \varepsilon_k
f\biggl(x_0+\varepsilon_k &y,
v_k(y_N),0,\ldots,0,\frac1{\varepsilon_k}v'_k(y_N)\biggr)
\le f^\infty (x_0+\varepsilon_k y, v_k(y_N),0,\ldots,0,v'_k(y_N))\\&+
\varepsilon_k f(x_0+\varepsilon_k y,  v_k(y_N),0)
\le f^\infty (x_0+\varepsilon_k y, v_k(y_N),0,\ldots,0,v'_k(y_N) )+
C\,\varepsilon_k.\endaligned$$
Therefore, Fubini's Theorem yields
$$\aligned\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
&\le\liminf_{k\to \infty}\int_{Q}
  f^\infty (x_0+\varepsilon_k y, v_k(y_N),0,\ldots,0,v'_k(y_N))\,dy
\\&=\liminf_{k\to \infty}n_k
\int_{Q'}\left(\int_{-1/2n_k}^{1/2n_k}
 f^\infty(x_0+\varepsilon_k y, v_k(y_N),
(u^+(x_0)-u^-(x_0)) \,e_N)\,dy_N\right)\,dy',
\endaligned\tag3.3$$
where $Q'$ is the unit cube in $\Bbb R^{N-1}$, and where we have used the fact that
$f^\infty$ is
positively homogeneous in $\xi$. Following \cite{5}, we  introduce now the Yosida transforms
$$f_\lambda(x,u,\xi):=\sup\bigl\{f^\infty(x',u,\xi)-\lambda|x'-x|:
\, x'\in\Omega\bigr\}, \tag3.4$$
for $\lambda>0$.  For $\lambda\le\eta$ and  by (1.8) it follows that
$$0\le f^\infty(x,u,\xi)\le f_\eta(x,u,\xi)\le f_\lambda(x,u,\xi)\le C|\xi|.
\tag3.5$$
We claim that
$$\lim_{\lambda\to\infty}f_\lambda(x,u,\xi)=f^\infty(x,u,\xi).\tag3.6$$
Indeed, let $\lambda>1$ and choose $x_\lambda$  such that
$$f_\lambda(x,u,\xi)\le f^\infty(x_\lambda,u,\xi)
-\lambda|x_\lambda-x|+\frac1\lambda.$$
By (3.5)
$$\aligned
f^\infty(x,u,\xi)\le f_\lambda(x,u,\xi)\le f_\lambda(x,u,\xi)
+\lambda|x_\lambda-x|\le f^\infty(x_\lambda,u,\xi)
+\frac1\lambda.
\endaligned\tag3.7$$
Since the right hand side is bounded by $C|\xi|+1$ and $f_\lambda\ge 0$,
it follows that $x_\lambda\to x$ as $\lambda\to\infty$.
If we now let $\lambda\to\infty$ in (3.7), and
use the fact that $f^\infty(\cdot,u,\xi)$ is upper semicontinuous,
we obtain (3.6).\par
Next we show that $f_\lambda$ is Lipschitzian. Fix $\varepsilon>0$,
$x$, $x_1\in\Omega$, and find
$x_\varepsilon$ such that
$$\aligned f_\lambda(x,u,\xi)&\le f^\infty(x_\varepsilon,u,\xi)
-\lambda|x_\varepsilon-x|+\varepsilon\\&\le
f^\infty(x_\varepsilon,u,\xi)
-\lambda|x_\varepsilon-x_1|+\lambda|x-x_1|+\varepsilon\le
f_\lambda(x_1,u,\xi)+\lambda|x-x_1|+\varepsilon.\endaligned$$
If we now let $\varepsilon\to 0$ we obtain
$$f_\lambda(x,u,\xi)-f_\lambda(x_1,u,\xi)\le \lambda|x-x_1|, $$
and, in a similar way,
$$f_\lambda(x_1,u,\xi)-f_\lambda(x,u,\xi)\le \lambda|x-x_1|.$$
We conclude that
$$
|f_\lambda(x,u,\xi)-f_\lambda(x_1,u,\xi)|\le \lambda|x-x_1|
\tag3.8
$$
for all $x, x_1 \in \Omega, u \in \Bbb R, \xi \in \Bbb R^N$.
\par
Fix $\lambda>0$. By (3.3), (3.5), and the fact that $f_\lambda$ is Lipschitzian, we have
$$\aligned\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
&\le\liminf_{k\to \infty}n_k
\int_{Q'}\left(\int_{-1/2n_k}^{1/2n_k}
 f^\infty(x_0+\varepsilon_k y, v_k(y_N),
(u^+(x_0)-u^-(x_0)) \,e_N)\,dy_N\right)\,dy'\\
&\le\liminf_{k\to \infty}n_k
\int_{Q'}\left(\int_{-1/2n_k}^{1/2n_k}
 f_\lambda(x_0+\varepsilon_k y, v_k(y_N),
(u^+(x_0)-u^-(x_0)) \,e_N)\,dy_N\right)\,dy'\\
&\le
\liminf_{k\to \infty}n_k
\int_{Q'}\left(\int_{-1/2n_k}^{1/2n_k}
 f_\lambda(x_0, v_k(y_N),
(u^+(x_0)-u^-(x_0)) \,e_N)+\lambda\,\varepsilon_k|y|\,dy_N\right)\,dy'\\
&=\liminf_{k\to \infty}n_k\int_{-1/2n_k}^{1/2n_k}
 f_\lambda(x_0, v_k(y_N),
(u^+(x_0)-u^-(x_0)) \,e_N)\,dy_N.
\endaligned$$
A simple change of variables now yields
$$ \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\le
\frac1{u^+(x_0)-u^-(x_0)}\int_{u^-(x_0)}^{u^+(x_0)}
 f_\lambda(x_0, s,(u^+(x_0)-u^-(x_0))\,e_N)\,ds.$$
Letting $\lambda\to\infty$, by (3.5), (3.6), and Lebesgue Dominated
Convergence Theorem, we obtain
$$ \aligned\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)&\le
\lim_{\lambda\to\infty}\frac1{u^+(x_0)-u^-(x_0)}\int_{u^-(x_0)}^{u^+(x_0)}
 f_\lambda(x_0, s,(u^+(x_0)-u^-(x_0))e_N)\,ds\\&=
\int_{u^-(x_0)}^{u^+(x_0)}
 f^\infty(x_0, s,e_N)\,ds,\endaligned$$
where we have used again the fact that
$f^\infty$ is
positively homogeneous in $\xi$.
\hfill\qquad\qed


\bigskip
\noindent {\bf\S4. Proof of Theorem 1.5.}
\medskip
(i) {\it  (Lebesgue part)} We claim that
$g(x):=f(x,u(x),\nabla u(x))\in L^1\loc(\Omega;\Bbb R)$. To show this it is
clearly enough to prove that
$$\int_{Q(x_0,\delta)}g(x)\,dx<\infty$$
for any $x_0\in\Omega$ and $\delta>0$ sufficiently small.
Let $\delta>0$ correspond to $\varepsilon=1/2$  in (1.9). Then
$$\aligned
\infty>\lim_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx\ge\frac23
\liminf_{n\to\infty}
\int_{Q(x_0,\delta)} f(x_0,u_n(x),\nabla u_n(x))\,dx
-\frac13\,\delta^N.\endaligned$$
The functional
$$F_0(v,A):=\int_{A}f(x_0,v,\nabla v)\,dx
 +\int_{A}
f^\infty(x_0,v,dC(v))
+ \int_{S(v)\cap A}\left(
\int_{v^-(x)}^{v^+(x)}f^\infty(x_0,s,\nu_u)\,ds\right)d\Cal H^{N-1}$$
satisfies all the conditions of Theorem 2 in  \cite{12}, and
 $F_0(u_n,A)=\int_{A} f(x_0,u_n(x),\nabla u_n(x))\,dx$
since $u_n\in W^{1,1}\loc(\Omega;\Bbb R)$; thus,
taking $A:=Q(x_0,\delta)$,
$$\aligned
&\infty>\lim_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx\ge \frac23 \,F_0(u,Q(x_0,\delta))
-\frac13\delta^N
\ge \frac13 \int_{Q(x_0,\delta)}g(x)\,dx -\frac23\delta^N,
\endaligned\tag4.1$$
where we have used (1.9) a second time and $H$ is defined in (1.2). Therefore the claim is proved.\par
Let $\Omega(u)$ be the set of Lebesgue points of
$g$. Since $g\in L^1\loc(\Omega;\Bbb R)$, we have that
$\Cal L^N(\Omega\backslash \Omega(u))=0$.
We now  proceed essentially as in the proof of
 Theorem 1.1, starting from (2.4) up to (2.6), where without loss of
 generality, we may also assume that
$x_0\in  \Omega(u)$.
Fix $\varepsilon>0$ and let $\delta>0$ be such that (1.9) holds.
Choose $\varepsilon_k\to 0^+$ such that $\mu(\partial
Q(x_0,\varepsilon_k))=0$,  $\varepsilon_k\le\delta$, and
$$\aligned
\frac{d\mu}{d\Cal L^N}(x_0)&=\lim_{k\to \infty}\lim_{n\to \infty}\frac{1}{\varepsilon_k^N}\int_{Q(x_0,\varepsilon_k)}
f(x,u_n,\nabla u_n)\,dx
\\&\ge\frac1{1+\varepsilon}\liminf_{k\to \infty}\liminf_{n\to \infty}\frac{1}{\varepsilon_k^N}
\int_{Q(x_0,\varepsilon_k)}
f(x_0,u_n,\nabla
u_n)\,dx-\frac\varepsilon{1+\varepsilon}.\endaligned\tag4.2$$
For fixed $k$, again by Theorem 2 in \cite{12} applied this time
to  $F_0(\cdot,Q(x_0,\varepsilon_k))$,
we obtain
$$\liminf_{n\to \infty}\frac{1}{\varepsilon_k^N}\int_{Q(x_0,\varepsilon_k)}
f(x_0,u_n(x),\nabla u_n(x))\,dx\ge\frac{1}{\varepsilon_k^N} \int_{Q(x_0,\varepsilon_k)}
f(x_0,u(x),\nabla u(x))\,dx$$
(recall that  $f^\infty\ge 0$)
and, consequently,
$$\aligned
\frac{d\mu}{d\Cal L^N}(x_0)&
\ge\frac1{1+\varepsilon}\liminf_{k\to \infty}\frac{1}{\varepsilon_k^N}\int_{Q(x_0,\varepsilon_k)}
f(x_0,u,\nabla u)\,dx-\frac\varepsilon{1+\varepsilon}\\
&\ge\frac{1-\varepsilon}{1+\varepsilon}\liminf_{k\to \infty}\frac{1}{\varepsilon_k^N}\int_{Q(x_0,\varepsilon_k)}
f(x,u,\nabla u)\,dx-\frac{2\varepsilon}{1+\varepsilon}=
\frac{1-\varepsilon}{1+\varepsilon}\,f(x_0,u(x_0),\nabla u(x_0))-\frac{2\varepsilon}{1+\varepsilon}
\endaligned
$$
where we have used (1.9) and the fact that $x_0\in \Omega(u)$.
We  now let $\varepsilon\to 0^+$.\hfill\qquad\qed

\medskip
(ii){\it (Cantor part)} By (4.1) the function
 $h(x):=f^\infty\left(x,u(x),\frac{dC(u)}{d|C(u)|}(x)\right)
\in L^1\loc(\Omega;|C(u)|)$; thus, by Lebesgue--Besicovitch Differentiation
Theorem,
$$\lim_{\varepsilon\to 0^+}
\frac1{|C(u)|(Q_\nu(x_0,\varepsilon))}\int_{Q_\nu(x_0,\varepsilon)}h\,d|C(u)|=
h(x_0)$$
for $|C(u)|$ a.e. $x_0\in \Omega$.
Moreover, it is known that
for $C(u)$ a.e. $x_0\in \Omega$
$$\lim_{\varepsilon\to 0^+}
\frac{|C(u)|(Q_\nu(x_0,\varepsilon))}{|Du|(Q_\nu(x_0,\varepsilon))}=1;$$
hence,
$$\lim_{\varepsilon\to 0^+}
\frac1{|Du|(Q_\nu(x_0,\varepsilon))}\int_{Q_\nu(x_0,\varepsilon)}h(x)\,d|C(u)|(x)=
h(x_0)\tag4.3$$
for $|C(u)|$ a.e. $x_0\in \Omega$. Let $M_1(u)$ be the
set of all points of $\Omega$ which satisfy (4.3).
Then
$|C(u)|(\Omega\backslash M_1(u))=0$.
We now proceed as in the proof of part (ii) in Theorem 1.1 up to (2.10),
with the only difference that we impose the further restriction that
$x_0\in  M_1(u)$, to obtain
$$
\frac{d\mu}{d|C(u)|}(x_0)=
\lim_{k\to \infty}\lim_{n\to \infty}\frac1{|Du|(Q_\nu(x_0,\varepsilon_k))}
\int_{Q_\nu(x_0,\varepsilon_k)}
f(x,u_n,\nabla u_n)dx.$$
The remaining of the proof follows an argument similar to that of
part (ii) above  after (4.2),  except that the integral is now averaged over
$Q_\nu(x_0,\varepsilon_k)$, and we use (4.3) at the end. We omit the details.
\hfill\qquad\qed

\medskip
(iii){\it  (Jump part)}
 It suffices to use arguments similar to those of parts (i) and
 (ii) above.
\hfill\qquad\qed



\bigskip
\noindent {\bf\S5. Proof of Theorem 1.6.}
\medskip
(i) {\it (Lebesgue part)} Fix $n\in\Bbb N$.
Applying the Scorza--Dragoni Theorem to the function
$f:\Omega\times [-n,n]\times\Bbb R^N\to[0,\infty)$
for each $i\in\Bbb N$ there exists a compact set
$K_{i,n}\subset\Bbb R$, with $\Cal L^1([-n,n]\backslash K_{i,n})\le1/(i2^n)$, such that
$f:\Omega\times K_{i,n}\times\Bbb R^N\to[0,\infty)$ is continuous.
 Let $K_{i,n}^*$ be the
set of Lebesgue points of $\chi_{K_{i,n}}$, and set $\omega:=\cup_{n}^\infty
\cup_{i}^\infty (K_{i,n}\cap K_{i,n}^*)$.
 Then
$$\Cal L^1(\Bbb R\backslash\omega)\le \sum_{n=1}^\infty
\Cal L^1([-n,n]\backslash K_{i,n})\le\frac1i
\to 0\quad\text{as $i\to\infty$},$$
and so $|Du|(A)=0$, where
 $A:=\{x\in\Omega\backslash S(u): u(x)\in \Bbb R\backslash\omega)\}$.
 Fix $x_0\in \Omega\backslash S(u)$. If $x_0\in A$ then, up to a set
 of $N-$dimensional Lebesgue measure zero, we may assume that $\nabla
 u(x_0)=0$, so that, as in the proof of Theorem 1.3(i), we have
$$f_1(x_0,u(x_0),0) = \lim_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\,\Cal F(u(x_0),Q(x_0,\varepsilon))\le
\limsup_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\int_{Q(x_0,\varepsilon)} f(x,u(x_0),0)\,dx=f(x_0,u(x_0),0),$$
where we have used the fact that $f(\cdot,u(x_0),0)$ is continuous.
 If $x_0\in (\Omega\backslash S(u))\backslash A $ and  $\nabla
 u(x_0)=0$, then we proceed as above.
 If $x_0\in (\Omega\backslash S(u))\backslash A $ and  $\nabla
 u(x_0)\ne 0$, then set
$$\nu:=\frac{\nabla u(x_0)}{|\nabla u(x_0)|},\qquad
\xi:=\nabla u(x_0),\qquad w_0(x):= u(x_0)+\xi(x-x_0).$$
Find
 $n\in\Bbb N$ such that $||w_0||_{L^\infty(Q_\nu(x_0,1))}\le n$ and
let  $i\in \Bbb N$ be such that   $u(x_0)\in K_{i,n}\cap K_{i,n}^*$.
Since $g(x):=f(x,w_0(x),\xi)$ is continuous over $ w_0^{-1}(K_{i,n})$, given $\delta>0$
there exists $\eta>0$ such that
$g(x)\le g(x_0)+\delta$ for all $x\in w_0^{-1}(K_{i,n})$ with $|x-x_0|\le\eta$.
Therefore, as in proof of Theorem 1.3(i), and by (1.8),
$$\aligned f_1(x_0,u(x_0),\xi)-|\xi|&
=   \lim_{\varepsilon\to 0^+}
\frac1{\varepsilon^N}\,\Cal F(u(x_0),Q(x_0,\varepsilon))\\
&\le (f(x_0,u(x_0),\xi)+\delta)\,\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q_\nu(x_0,\varepsilon)\cap w_0^{-1}(K_{i,n}))}{\varepsilon^N}\\
&\quad +C(1+|\xi|)
\,\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q_\nu(x_0,\varepsilon)\backslash w_0^{-1}(K_{i,n}))}{\varepsilon^N}\\
& \leq f(x_0,u(x_0),\xi)+\delta,
\endaligned\tag5.1$$
since, by the coarea formula (see \cite{31, Theorem 2.7.1}), and setting $\eta_{\varepsilon}(x) := u(x_0) + \varepsilon \xi\cdot x$ for $x \in Q_\nu$, we have
$$
\aligned
\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q_\nu(x_0,\varepsilon)\backslash w_0^{-1}(K_{i,n}))}{\varepsilon^N} & =
1 - \liminf_{\varepsilon\to 0^+}\frac{\Cal L^N(Q_\nu(x_0,\varepsilon)\cap w_0^{-1}(K_{i,n}))}{\varepsilon^N}\\
&=1 -\liminf_{\varepsilon\to 0^+} \frac1{|\xi|\,\varepsilon^N}
\int_{Q_\nu(x_0,\varepsilon)\cap w_0^{-1}(K_{i,n})}
|D w_0(x)|\,dx\\
&
=1 - \liminf_{\varepsilon\to 0^+}\frac1{|\xi|\,\varepsilon^N}\int_{K_{i,n}}
\Cal H^{N-1}(w_0^{-1}(s)\cap Q_\nu(x_0,\varepsilon))\, ds.
\endaligned
$$
For simplicity, we assume that
$x_0=0$, $u(x_0)=0$, $\nu=e_N$,
so that
$$w_0^{-1}(s)\cap Q_\nu(x_0,\varepsilon)=
\cases \varepsilon[-\frac12,\frac12]^{N-1}\times\{s\}\quad&
\text{if $|s|\le\frac\varepsilon2\,|\xi|,$}\\
\emptyset \quad&\text{if $|s|>\frac\varepsilon2\,|\xi|;$}\endcases$$
thus,
$$
\limsup_{\varepsilon\to 0^+}
\frac{\Cal L^N(Q_\nu(x_0,\varepsilon)\backslash w_0^{-1}(K_{i,n}))}{\varepsilon^N}=
1-\liminf_{\varepsilon\to 0^+}\frac{\Cal L^1([-\frac\varepsilon2\,|\xi|,\frac\varepsilon2\,|\xi|]\cap  K_{i,n})}
{|\xi|\varepsilon} = 0
$$
where we have used the fact that $u(x_0)$ is a  Lebesgue point of
$\chi_{K_{i,n}}$.
By letting $\delta\to 0^+$ in (5.1) we obtain the desired inequality.
\hfill\qquad\qed

\medskip
(ii) {\it (Cantor part)}
\par
The proof for the Cantor part  is very similar to the
previous one (see also the proof of Theorem 1.3(ii)), and therefore we
omit the details.
\hfill\qquad\qed


\medskip
(iii) {\it (Jump part)}
The proof follows
the same arguments of the proof of Theorem 1.3(iii). \hfill\qquad\qed

\bigskip
\noindent {\bf\S6. Proof of Theorem 1.7.}
\medskip
(i) {\it (Lebesgue part)}
We proceed as in the proof of Theorem 1.1(i) up to (2.7), where instead of
using the truncated sequence, we apply condition (1.10) to get
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+
\varepsilon\ge\liminf_{k\to\infty}
\int_{Q}
f(x_0,u(x_0),\nabla w_k(y))\,dy-\int_{Q}\rho(\varepsilon_k\,|w_k(y)|)\,dy.$$
By Fatou's Lemma, and since $\rho$ is continuous with $\rho(0) = 0$, we have
$$
\aligned
C - \limsup_{k\to\infty}\int_{Q}\rho(\varepsilon_k\,|w_k(y)|)\,dy &=
\liminf_{k\to\infty}\int_{Q}[C(1 +\varepsilon_k\,|w_k(y)|) - \rho(\varepsilon_k\,|w_k(y)|)]\,dy\\
&\geq \int_Q \liminf_{k\to\infty}[C(1 +\varepsilon_k\,|w_k(y)|) - \rho(\varepsilon_k\,|w_k(y)|)]\,dy = C,
\endaligned
$$
and so
$$
 \int_{Q}\rho(\varepsilon_k\,|w_k(y)|)\,dy \to 0\quad \text{as $k\to\infty$}.$$
Thus
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+
\varepsilon\ge\liminf_{k\to\infty}
\int_{Q}
f(x_0,u(x_0),\nabla w_k(y))\,dy.\tag6.1$$
If $g(\xi):= f(x_0,u(x_0),\xi)$ is convex then
we may  apply Serrin's Theorem A(i),
which continues to hold in the vectorial case.  If
$g$ is quasiconvex and $q=1$ in (1.11), then we apply a result of
 Ambrosio and Dal Maso \cite{4} (see also \cite{18})
 to conclude that
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon\ge
f(x_0,u(x_0),\nabla u(x_0)).$$
It is now sufficient to let $\varepsilon \to 0^+$ to obtain the desired result.
 When $q>1$ and $g$ is quasiconvex in (1.11), then we can apply an approximation result of
Kristensen \cite{22, Proposition 1.9} to write
$$f(x_0,u(x_0),\xi)=\sup_{j}g_j(\xi),$$
where $g_j(\xi)$ is quasiconvex, $g_j(\xi)\le g_{j+1}(\xi)$,
and $g_j(\xi)=a_j|\xi|+b_j$ for $|\xi|$ large, say $|\xi|\ge r_j$. From
(6.1) and
for any fixed $j$, applying \cite{4} we have
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+
\varepsilon\ge\liminf_{k\to\infty}
\int_{Q}
g_j(\nabla w_k(y))\,dy\ge
g_j(\nabla u(x_0)),$$
and then let $j\to\infty$.
\hfill\qquad\qed
\medskip
(ii) {\it (Cantor part)} We proceed as in Theorem 1.1(ii)
until (2.11), where (2.8) should  now be written as
$$\frac{d C(u)}{d|C(u)|}(x_0)=a_u(x_0)\otimes\nu_u(x_0),\tag6.2$$
with $a_u(x_0)\in\Bbb R^d$
and $\nu_u(x_0)\in S^{N-1}$, and where we have used Alberti's result
 \cite{2}. By (2.11) and (1.10)
$$\aligned
(1+\varepsilon)\frac{d\mu}{d|C(u)|}(x_0)+\varepsilon
&\ge
\lim_{k\to \infty}\left\{
\frac1{t_k}\int_{Q_\nu}f(x_0,u(x_0),t_k\nabla w_k(y))\,dy\right.\
\\&\quad -\left.\frac1{t_k}\int_{Q_\nu}\rho(|u(x_0)-u_{0k}+\lambda_k w_k(y)|)\,dy\right\}.
\endaligned\tag6.3
$$
As $\rho(s) \leq C(1 + s)$ for all $s > 0$ and for some $C > 0$, $w_k$  converges to $w_0$ in $L^1(Q_\nu;\Bbb R^d)$,
$u_{0k}$ converges to $u(x_0)$,
$\lambda_k\to 0$, and
$t_k\to \infty$ as $k\to\infty$, we have
$$
\frac1{t_k}
\int_{Q_\nu}\rho(|u(x_0)-u_{0k}+\lambda_k\,w_k(y)|)\,dy \rightarrow 0
$$
and thus
$$\aligned
(1+\varepsilon)\frac{d\mu}{d|C(u)|}(x_0)+\varepsilon\ge
\lim_{k\to \infty}
\frac1{t_k}\int_{Q_\nu}&f(x_0,u(x_0),t_k\nabla w_k(y))\,dy.
\endaligned\tag6.5
$$
If
$g$ is quasiconvex and $q=1$ in (1.11) then we can proceed as in Theorem 1.1(ii)
starting from (2.12).
When  $g(\xi):= f(x_0,u(x_0),\xi)$ is convex, or $g$ is quasiconvex
$q>1$ in (1.11),
then we use Proposition 9.1 below  or Proposition 1.9 of
Kristensen \cite{22}, to deduce from (6.3) and (6.4):
$$
(1+\varepsilon)\frac{d\mu}{d|C(u)|}(x_0)+\varepsilon
\ge
\lim_{k\to \infty}
\frac1{t_k}\int_{Q_\nu}g_j(t_k\nabla w_k(y))\,dy.
$$
Proceeding as in the case $q=1$, we obtain
$$\frac{d\mu}{d|C(u)|}(x_0)\ge
g_j^\infty(a\otimes \nu).$$
Since the function $h_j(t):= g_j(t a\otimes \nu)$ is convex, by
Proposition 9.1, (9.1),  we have for $t>1$
$$\frac{d\mu}{d|C(u)|}(x_0)\ge
\frac{g_j(t\,a\otimes \nu)}t-\frac{g_j(0)}t,\qquad t>1.$$
As $f(x_0,u(x_0),0)=\sup_j g_j(0)\le C$,  letting $j\to\infty$
yields
$$\frac{d\mu}{d|C(u)|}(x_0)\ge\frac{f(x_0,u(x_0),t\,a\otimes \nu)}t
-\frac{C}t.$$
We now let $t\to\infty$.
\hfill\qquad\qed
\medskip
\remark{Remark } Note that when (1.12) holds then
$$f^\infty(x_0,u_0,\xi)=\cases +\infty\quad&\text{if $\xi\ne 0$}\\
                                0 \quad&\text{if $\xi= 0$.}\endcases$$
\endremark


\medskip
(iii){\it (Jump part)} We proceed as in Theorem 1.1(iii)
up to  (2.14). By (2.14) and (1.10)
$$
(1+\varepsilon)\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
+\varepsilon
\ge
\lim_{k\to \infty}
\int_{Q_\nu}\varepsilon_k f\left(x_0,\frac1{\varepsilon_k}\nabla w_k(y)\right)\,dy
$$
Now we continue exactly as in the proof of the Cantor part in
Theorem 1.7, starting from
(6.5), with  the vector $a$ in place of $u^+(x_0)-u^-(x_0)$.
\hfill\qquad\qed




\bigskip

\noindent {\bf\S7. Proof of Theorems 1.8--1.10.}
\medskip
\demo{Proof of Theorem 1.8 (Lebesgue part)}
We proceed as in Theorem 1.1(i) until (2.7). If $f(x_0,u(x_0),\xi)\equiv 0$
for all $\xi$ then there is nothing to prove. Thus, we assume
that (1.13) and (1.14) hold, we fix $\varepsilon>0$, and
 let $\delta>0$ be given by (1.13) and (1.14). \par
{\it Step 1.} We   prove first the theorem under the additional hypothesis that there exists $M >0$ such that
$$0\le f(x,u,\xi)\le M(1+\xi)\tag7.1$$
for all $x\in\Omega$ with $|x-x_0|\le\delta$, $u\in\Bbb R^d$, and
$\xi\in\Bbb R^{dN}$.
As in \cite{17, Proposition 2.6 }, we may find
$w_k\in C_0^\infty(\Bbb R^N;\Bbb R^d)$, with $w_k\to w_0$ in $L^1$,
 $w_0(x):=\nabla u(x_0)\,x$, such that by (2.7), (1.14), and
for $k$ sufficiently large,
$$
\frac{d\mu}{d\Cal L^N}(x_0)\ge \int_{Q\cap\{|w_k|\le\delta/\varepsilon_k\}}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kw_k(y),\nabla w_k(y))\,dy\ge
C_1\int_{Q\cap\{|w_k|\le\delta/\varepsilon_k\}}|\nabla w_k(y)|\,dy-C_2;
$$
 thus there exists a constant $K>0$ such that
$$\int_{Q\cap\{|w_k|\le\delta/\varepsilon_k\}}|\nabla w_k(y))|\,dy\le K.\tag7.2$$
In order to truncate $w_k$, fix $s_k>||w_0||_{L^\infty(Q;\Bbb R^d)}+1$,
 $L_k>s_k$,
 and construct
a smooth  cut--off function $g_k:\Bbb R^d\to \Bbb R^d$ such that
$$g_k(u)=\cases u\quad&\text{if $|u|\le s_k,$}\\
              0\quad&\text{if $|u|\ge L_k,$}\endcases$$
with $|g_k(u)|\le |u|$  and
$|D g_k(u)|\le C\,L_k/(L_k-s_k)$ for all $u\in\Bbb R^d$.
Define $v_k(y):=g_k(w_k(y))$,   and
$$ \aligned
E_k:=\{y\in Q: \,|w_k(y)|< s_k\},\quad
E_k^+:=\{y\in Q: \,|w_k(y)|> L_k \},\quad
E_k^-:=\{y\in Q: \,s_k\le |w_k(y)|\le L_k\}.\endaligned $$
    Then
$$\aligned
&\int_Q|v_k(y)-w_0(y)|dy=
\int_{E_k}|w_k(y)-w_0(y)|dy+
\int_{E_k^+}|w_0(y)|dy
+\int_{E_k^-}|g_k(w_k(y))-w_0(y)|dy\\
&\le
||w_k-w_0||_{L^1(Q;\Bbb R^d)}+||w_0||_{L^\infty(Q;\Bbb R^d)}\Cal L^N(E_k^-
\cup E_k^+)+
\int_{E_k^-}|w_k(y)|dy\\
&\le 2||w_k-w_0||_{L^1(Q;\Bbb R^d)}
+2||w_0||_{L^\infty(Q;\Bbb R^d)}\Cal L^N(E_k^-\cup E_k^+)
\to 0\qquad\text{as $k\to\infty$,}\endaligned$$
 because
$$\aligned 0\le \Cal L^N(E_k^-\cup E_k^+)
=& \Cal L^N(\{y\in Q: \, |w_k(y)|\ge s_k\})\\
&\le
\Cal L^N(\{y\in Q: \, |w_k(y)-w_0(y)|\ge 1\})\le ||w_k-w_0||_{L^1(Q;\Bbb R^d)}.
\endaligned\tag7.3$$
Moreover
$$\aligned
&\int_{Q}f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_k v_k(y),\nabla v_k(y))\,dy
=\int_{E_k}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_k w_k(y),\nabla w_k(y))\,dy\\
&+
\int_{E_k^+}
f(x_0+\varepsilon_k y,u(x_0),0)\,dy+
\int_{E_k^-}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_k v_k(y),\nabla v_k(y))\,dy.
\endaligned\tag7.4$$
We claim that the last two integrals  are
infinitesimal as $k\to\infty$.
Indeed, by (7.1) and (7.3),
$$0\le \int_{E_k^+}
f(x_0+\varepsilon_k y,u(x_0),0)\,dy\le M\Cal L^N(E_k^+)\to 0,$$
while, from (7.1) and the coarea formula,
$$\aligned
&\int_{E_k^-}f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_k v_k(y),\nabla v_k(y))\,dy
\le M\int_{E_k^-}\left(1+|\nabla g_k(w_k)\nabla w_k|\right)dy\\
&\le
M\left(\Cal L^N(E_k^-)+\frac{CL_k}{L_k-s_k}\int_{E_k^-}|\nabla
w_k|\,dy\right)\\
&=M\left(\Cal L^N(E_k^-)+\frac{CL_k}{L_k-s_k}\int_{s_k}^{L_k}\Cal H^{N-1}(\{y\in Q:\,
|w_k(y)|=t\})\,dt\right).
\endaligned\tag7.5$$
By Theorem 7.10 of \cite{28} and (7.2),
 for $\Cal L^1$ a.e. $L\le\delta/\varepsilon_k $ we have
$$\lim_{s\to L}\frac1{L-s}\int_s^L\Cal H^{N-1}(\{y\in Q:\,
|w_k(y)|=t\})\,dt= \Cal H^{N-1}(\{y\in Q:\, |w_k(y)|=L\}).\tag7.6$$
Moreover, by Lemma 2.6 in \cite{17} and (7.2),
 for any $0<\alpha<\beta<\delta/\varepsilon_k$ we obtain
$$\mathop{\text{\rm essinf}}_{L\in (\alpha,\beta)}L
\Cal H^{N-1}(\{y\in Q:\, |w_k(y)|=L\})\le \frac{K}{\log(\beta/\alpha)}.$$
Set $\alpha:=\delta/\varepsilon_k^{1/4}$ and
$\beta:=\delta/\varepsilon_k^{1/2}$,
and  find $L_k\in(\delta/\varepsilon_k^{1/4},\delta/\varepsilon_k^{1/2})$ such that
(7.6) holds and
$$L_k
\Cal H^{N-1}(\{y\in Q:\, |w_k(y)|=L_k\})\le
\frac{2K}{\log(1/\varepsilon_k^{1/4})}.$$
Choose $s_k\ge L_k/2$ so that
$$\frac{L_k}{L_k-s_k}\int_{s_k}^{L_k}\Cal H^{N-1}(\{y\in Q:\,
|w_k(y)|=t\})\,dt\le \frac{2K}{\log\left(1/\varepsilon_k^{1/4}\right)}+\frac1k.$$
Then the integral in the right hand side of (7.5) approaches zero
 as $k\to\infty$, and so,  from (7.4),
$$\frac{d\mu}{d\Cal L^N}(x_0)\ge\liminf_{k\to\infty}
\int_{Q}
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kv_k(y),\nabla v_k(y))\,dy.$$
Since $\varepsilon_k\to 0$, by (1.13) we obtain
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon\ge\liminf_{k\to\infty}
\int_{Q}
f(x_0,u(x_0),\nabla v_k(y))\,dy.$$
We can now continue as in the proof of Theorem 1.7(i),  and
the result is established
if $f$ satisfies (7.1).\par
{\it Step 2.} In the general case, let
$\psi\in C^\infty_0(\Bbb R^d;\Bbb R)$ be a cut--off function,
with  $0\le\psi\le 1$,
and such that
$\psi\equiv 1$ on $B(u(x_0),\delta/2)$,
$\psi\equiv 0$ outside $B(u(x_0),\delta)$.
From (2.7)
$$
\frac{d\mu}{d\Cal L^N}(x_0)\ge\liminf_{k\to\infty}
\int_{Q}\psi(u(x_0)+\varepsilon_kw_k(y))
f(x_0+\varepsilon_k y,u(x_0)+\varepsilon_kw_k(y),\nabla w_k(y))\,dy,$$
and by (1.13),
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon
\ge\liminf_{k\to\infty}
\int_{Q}\psi(u(x_0)+\varepsilon_kw_k(y))
f(x_0,u(x_0),\nabla w_k(y))\,dy.$$
If $f(x_0,u(x_0),\cdot)$ is convex (resp. quasiconvex with $q>1$ in
(1.11)),
we  use Proposition 9.1  (resp. Proposition 1.9 of
Kristensen \cite{22})
 to  approximate $f(x_0,u(x_0),\xi)$ by an increasing
sequence $g_j(\xi)$ of convex (resp. quasiconvex) functions such that
$$0\le g_j(\xi)\le C_j(|\xi|+1).\tag7.7$$
If $f(x_0,u(x_0),\cdot)$ is  quasiconvex with $q=1$ in (1.11),
we simply take $g_j(\xi)\equiv f(x_0,u(x_0),\xi) $ for all $j$.
For any fixed $j$
$$(1+\varepsilon)\frac{d\mu}{d\Cal L^N}(x_0)+\varepsilon
\ge\liminf_{k\to\infty}
\int_{Q}\psi(u(x_0)+\varepsilon_kw_k(y))
g_j(\nabla w_k(y))\,dy,$$
and
$f_j(x,u,\xi):=\psi(u)g_j(\xi)$ satisfies (7.1). Moreover, by (7.7),
(7.2) continues to hold, provided we replace $\delta$
with $\delta/2$. Finally, (1.13) is still satisfied {\it at the point }
$(x_0,u(x_0))$. Therefore we can apply the first part of the proof to get
$$\frac{d\mu}{d\Cal L^N}(x_0)\ge \psi(u(x_0))
g_j(\nabla u(x_0))=g_j(\nabla u(x_0)).$$
It suffices to take the supremum in $j$.
\hfill\qquad\qed
\enddemo
\medskip
\demo{Proof of Theorem 1.8 (Cantor part)}
We proceed as in Theorem 1.1(ii) until (2.11). We can now truncate the
sequence $w_k$ using an argument similar to that of the Lebesgue part of Theorem 1.8
(note that the only property of $w_0(y)$
 which has been used is the fact that it is bounded), and then continue
as in the Cantor part of Theorem 1.7, using (1.13) in place
of (1.11). We omit the details.
\hfill\qquad\qed
\enddemo
\demo{Proof of Theorem 1.9} The proofs of Theorem 1.3(i)--(ii) and of the
first part of Theorem
1.6 continue to hold.
We  observe that in Theorem 1.3(ii),  since $a\otimes\nu$ has rank
one,
the function $g(t)=f^\infty(x,u,t\,a\otimes\nu)$ is convex and thus
we can still use  Proposition 9.1, (9.1).
\par
If $f^\infty=f^\infty(x,\xi)$  then the proof of Theorem
1.3(iii) is still valid with some obvious modifications.
\hfill\qquad\qed
\enddemo
\demo{Proof of Theorem 1.10} We proceed as in Theorem 1.1(iii) until (2.14).
 Fix $\varepsilon>0$ and let $k$ be so large that
$\varepsilon_k<\min\{\delta,1/L\}$, where $\delta$ and $L$ are
provided by (1.15)--(1.16). Then by (1.16) and (1.15), in this order,
$$\aligned
\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)&=
\lim_{k\to \infty}\int_{Q_\nu}
\varepsilon_k f\left(x_0+\varepsilon_k y, w_k(y),\frac1{\varepsilon_k}
\nabla w_k(y)\right)\,dy
\\&
\ge\frac1{1+\varepsilon}
\liminf_{k\to \infty}\int_{Q_\nu}
 f^\infty(x_0+\varepsilon_k y, w_k(y),\nabla w_k(y))\,dy
-\frac\varepsilon{1+\varepsilon}\\
&\ge\frac1{(1+\varepsilon)^2}
\liminf_{k\to \infty}\int_{Q_\nu}
f^\infty(x_0, w_k(y),\nabla w_k(y))\,dy
-\frac{\varepsilon}{(1+\varepsilon)^2}-\frac{\varepsilon}{1+\varepsilon}\\
&\ge\frac1{(1+\varepsilon)^2}
\limsup_{k\to \infty}\int_{Q_\nu}
f^\infty(x_0, v_k(y),\nabla v_k(y))\,dy-\frac{\varepsilon}{(1+\varepsilon)^2}-\frac{\varepsilon}{1+\varepsilon},
\endaligned\tag7.8$$
where we have  used Lemma 2.6 and Remark 2.7(1) of \cite{7} to obtain
 a new sequence
$v_k \in W^{1,1}(Q_\nu;\Bbb R^d)$ which converges to $w_0$ in $L^1(Q_\nu;\Bbb R^d)$ and
 such that $ v|_{\partial Q_\nu}=w_0$. It now follows from (7.8)
 and the definition of the function $h$ in (1.18) that
$$
\frac{d\mu}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\ge
\frac1{(1+\varepsilon)^2}\,h(x_0,u^+(x_0),u^-(x_0),\nu)
-\frac{\varepsilon}{(1+\varepsilon)^2}-\frac{\varepsilon}{1+\varepsilon},$$
and we obtain the first part of the
theorem upon letting $\varepsilon\to 0^+.$
\medskip
We first prove the reverse inequality to (1.17) under the additional coercivity assumption that there exists $C>0$ such that
$$ f(x,u,\xi)\ge C|\xi|\qquad\text{
 for all $(x,u,\xi)\in\Omega\times\Bbb R^d\times\Bbb R^{dN}$. }\tag7.9$$
 Fix $\varepsilon_0>0$, and define
$$u_0(x):=
\cases 
u^+(x_0)  &\quad\text{if $y\cdot\nu>0,$}\\
       u^-(x_0)&\quad\text{if $y\cdot\nu\le0.$}\endcases
$$
By Lemma 4.1.3 and (3.17) of \cite{7}
for
$\Cal H^{N-1}$ a.e. $x_0\in S(u)$
$$\aligned 
&\frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\\
& \quad =\limsup_{\varepsilon\to 0^+}
\frac{1}{\varepsilon^{N-1}}
\inf\left\{\int_{Q_\nu(x_0,\varepsilon)}\!\!\!\!f(x,v(x),\nabla v(y))dy:
v\in W^{1,1}(Q_\nu(x_0,\varepsilon);\Bbb R^d), v|_{\partial Q_\nu(x_0,\varepsilon)}
=u_0(\cdot - x_0) \right\}\\
&\quad =\limsup_{\varepsilon\to 0^+}
\,\inf\left\{\int_{Q_\nu}\varepsilon\,
f\left(x_0+\varepsilon\,y,w(y),\frac1\varepsilon\nabla w(y)\right)\,dy:\,
w\in W^{1,1}(Q_\nu;\Bbb R^d),\, w|_{\partial Q_\nu}
=u_0\right\}\\
&\quad  \le
\frac1{1-\varepsilon_0}\limsup_{\varepsilon\to 0^+}
\,\inf\left\{\int_{Q_\nu}
f^\infty(x_0+\varepsilon\,y,w(y),\nabla w(y))\,dy:\,
w\in W^{1,1}(Q_\nu;\Bbb R^d),\, w|_{\partial Q_\nu}
=u_0\right\} +\frac{\varepsilon_0}{1-\varepsilon_0}\\
&\quad \le
\frac1{1-\varepsilon_0}\limsup_{\varepsilon\to 0^+}\int_{Q_\nu}
f^\infty(x_0+\varepsilon\,y,w_1(y),\nabla w_1(y))\,dy
+\frac{\varepsilon_0}{1-\varepsilon_0},
\endaligned$$
for any
$w_1\in W^{1,1}(Q_\nu;\Bbb R^d)$, with  $w_1|_{\partial Q_\nu}
=u_0$, and
where  we have used $(1.16)^{\prime}$.
We now take $w_1$ in the previous inequality such that
$$
\int_{Q_\nu}f^\infty(x_0,w_1(y),\nabla w_1(y))\,dy\le
h(x_0,u^+(x_0),u^-(x_0), \nu_u(x_0))+\varepsilon_0.\tag7.10$$
By (3.5), (3.6), and Lebesgue Dominated
Convergence Theorem,
$$\lim_{\lambda\to\infty}\int_{Q_\nu}f_\lambda(x_0,w_1(y),\nabla w_1(y))\,dy=
\int_{Q_\nu}f^\infty(x_0,w_1(y),\nabla w_1(y))\,dy,\tag7.11$$
where the Yosida transforms $f_\lambda$ were introduced in (3.4);
 thus, for fixed $\lambda$ sufficiently large, by (7.10), and by (7.11)
$$
\int_{Q_\nu}f_\lambda(x_0,w_1(y),\nabla w_1(y))\,dy\le
h(x_0,u^+(x_0),u^-(x_0), \nu_u(x_0))+2\varepsilon_0.\tag7.12$$
Consequently, also from (3.5) and (3.8),
$$\aligned \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
\le
\frac1{1-\varepsilon_0}&\limsup_{\varepsilon\to 0^+}\int_{Q_\nu}
f_\lambda(x_0+\varepsilon\,y,w_1(y),\nabla w_1(y))\,dy
+\frac{\varepsilon_0}{1-\varepsilon_0}\\
\le
\frac1{1-\varepsilon_0}&\limsup_{\varepsilon\to 0^+}\left(\int_{Q_\nu}
f_\lambda(x_0,w_1(y),\nabla w_1(y))\,dy+\varepsilon\,\lambda\right)
+\frac{\varepsilon_0}{1-\varepsilon_0}\\
\le\frac1{1-\varepsilon_0}&\left(
h(x_0,u^+(x_0),u^-(x_0), \nu_u(x_0))+2\varepsilon_0
\right)
+\frac{\varepsilon_0}{1-\varepsilon_0}\endaligned$$
by (7.12). Letting $\varepsilon_0\to 0^+$ in the previous inequality
 yields the desired result when (7.9) holds. 

In the general case, it
 suffices to consider the family of perturbed energy densities
 $$f_\rho(x,u,\xi):=f(x,u,\xi)+\rho\,|\xi|,\qquad\rho>0.$$
Then, since $\Cal F\le\Cal F_\rho$, it follows that
$$\aligned & \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)
\le\frac{d\Cal F_\rho(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\\
&\quad \le\inf\left\{\int_{Q_\nu}f^\infty(x_0,w(y),\nabla w(y))\,dy+\rho
\int_{Q_\nu}|\nabla w(y))|\,dy :\,
w\in W^{1,1}(Q_\nu;\Bbb R^d),\, w|_{\partial Q_\nu(x_0,\varepsilon)}=u_0\right\}\\
&\le 
\int_{Q_\nu}f^\infty(x_0,w_1(y),\nabla w_1(y))\,dy+\rho
\int_{Q_\nu}|\nabla w_1(y))|\,dy
\endaligned$$
for any fixed $w_1\in W^{1,1}(Q_\nu;\Bbb R^d)$, such that
$w_1|_{\partial Q_\nu(x_0,\varepsilon)}=u_0$.
 Letting $\rho\to 0^+$ yields
$$ \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\le
\int_{Q_\nu}f^\infty(x_0,w_1(y),\nabla w_1(y))\,dy,$$
and since $w_1$ is arbitrary, by taking the infimum over all functions
$w_1$ we obtain
$$ \frac{d\Cal F(u,\cdot)}{d\Cal H^{N-1}\lfloor S(u)}(x_0)\le
h(x_0,u^+(x_0),u^-(x_0), \nu_u(x_0)).$$
\hfill\qquad\qed
\enddemo
\bigskip

\noindent {\bf\S8. Further results.}
\medskip
As in Theorems C and D of Dal Maso \cite{9}, some of our results continue
to hold if the regularity conditions on the integrand $f$ are required
everywhere except at most on ``small" sets. In order to establish the main
result of this section, Theorem 8.5, we prove first three lemmas.

\medskip
\noindent\proclaim{Lemma 8.1} Let $N_0$  be a Borel subset of $\Omega\times\Bbb R$,
with $\Cal H^N(N_0)=0$, let $u\in BV(\Omega;\Bbb R)$, and define
$$ A:=\{x\in\Omega\backslash (S(u)\cup M(u)):\, (x,u(x))\in N_0\},\quad
B:=\{x\in M(u):\, (x,u(x))\in N_0\},$$ where $M(u)$ is the support of
$C(u)$. Then
\smallskip
\item{\rm (i)} $\Cal L^N(A)=0$;
\item{\rm (ii)} $|C(u)|(B)=0$.
\endproclaim
\demo{Proof} {\rm (i)} If  $\Cal L^N(A)>0$, then by Corollary 1 in Sec. 2.4.1
of \cite{15}  we obtain
$$0=\Cal H^N(N_0)\ge \Cal H^N(\{(x,u(x)):\, x\in A\})\ge \Cal H^N(A)=\Cal L^N(A)>0,$$
which is clearly a contradiction.\par
{\rm (ii)} Let
$$ \aligned G_+&:=\{(x,s)\in \Omega\times\Bbb R:\,s<u^+(x)\},\\
G_{\pm}&:=\{(x,s)\in \Omega\times\Bbb R:\,u^-(x)\le s\le u^+(x)\}.
\endaligned$$
Then $\chi_{G_+}\in BV\loc(\Omega\times\Bbb R)$ (see \cite{25}), and for any Borel set $K\subset
\Omega\times\Bbb R$ and $D\subset\Omega$ we have
$$|D \chi_{G_+}|(K)=\Cal H^N(K\cap G_{\pm}),\qquad
\int_{D\times\Bbb R}|D \chi_{G_+}|=\int_D|\zeta(u)|,\tag8.1
$$
where $\zeta(u):=(Du,-\Cal L^N)$ (see \cite{9}, Lemma 2.2). Take $K:=B\times\Bbb R$.
If $x\in M(u)$ then $u^+(x)=u^-(x)$, and thus
$$|D \chi_{G_+}|(K)=\Cal H^N(\{(x,u(x)):\, x\in B\})\le \Cal H^N(N_0)=0.$$
In turn, by (8.1), $\int_B|\zeta(u)|=0$. Since $|\zeta(u)|$ coincides with
$|Du|$ on $M(u)$ (recall that $\Cal L^N(M(u))=0$) it follows that $|D
u|(B)=0$.
\hfill\qquad\qed
\enddemo
\medskip
The following Lemma is a generalization of Theorem 3 in Sec. 2.4.3 of
\cite{15}
\medskip
\noindent\proclaim{Lemma 8.2}  Let  $h\in L^1\loc(\Omega;\Bbb R)$,
let $\mu$ be a positive Radon measure,  and define
$$B_0:=\left\{x_0\in\Omega:\, \limsup_{\varepsilon\to 0^+}
 \frac1{\mu(Q_\nu(x_0,\varepsilon))}\int_{Q_\nu(x_0,\varepsilon)}|h(x)|
 \,dx>0\right\},\tag8.2$$
where in the limsup we consider only those $\varepsilon>0$ such that
$\mu(\partial Q_\nu(x,\varepsilon))=0$.
 Then $\mu_s(B_0)=0$, where $\mu:=\dfrac{d\mu}{d\Cal L^N}\,\Cal L^N+\mu_s$.
\endproclaim
\demo{Proof}   Without loss of generality,
we can assume that $\Omega$ is bounded, $h\ge 0$ and
 $h\in L^1(\Omega;\Bbb R)$. Given $\eta>0$ there exists $\delta>0$  such that
$$\int_U h(x)\,dx<\eta\qquad\text{whenever $\Cal L^N(U)<\delta$}.$$
Let $E $ be the support of
 the measure $\mu_s$. Since $\mu_s(X)=\mu_s(X\cap E )$, we consider 
 $B_0'=B_0\cap E$. Clearly 
$\Cal L^N(B_0')=0$, and 
$$B_0'=\bigcup_{r\in \Bbb Q^+}B_r,\quad\text{where}
\quad B_r:=\left\{x_0\in E :\, \limsup_{\varepsilon\to 0^+}
 \frac1{\mu(Q_\nu(x_0,\varepsilon))}\int_{Q_\nu(x_0,\varepsilon)}
h(x)\,dx>r\right\}.$$

We claim that $\mu(B_r)=0$, from what will follow that $\mu(B_0')=0$. Let  $U$ be an open set such that $ B_r\subset U$ and $\Cal L^N(U)<\delta$.
Fix  $\rho>0$ and consider
$$\aligned \Cal F_1^{\rho}:&=\left\{Q_\nu(x,\varepsilon): x\in B_r,
0<\varepsilon<\rho,\,
\mu(\partial Q_\nu(x,\varepsilon))=0,\,\overline{Q_\nu(x,\varepsilon)}\subset U,\,
\int_{Q_\nu(x,\varepsilon)}h(y) dy>r\mu(Q_\nu(x,\varepsilon))\right\}
\endaligned
$$
and consider the Borel sets
$$
U^{\rho}:= \bigcup\left\{Q_\nu(x,\varepsilon) \, : \, Q_\nu(x,\varepsilon) \in \Cal F_1^{\rho}\right\}, \quad U_0 := \bigcap_{\rho>0} U^{\rho}.
$$
Since $B_r \subset U_0$, it suffices to prove that $\mu(U_0)=0$. Fix a compact set $K \subset U_0$, $\rho_0>0$, and let
$$
\aligned
\Cal F_2^{\rho_0}:&=\left\{Q_\nu(x,\varepsilon):\, x\in B_r,\, 0<\varepsilon<\rho,\,
\mu(\partial Q_\nu(x,\varepsilon))=0,\,\overline{Q_\nu(x,\varepsilon)}\subset U^{\rho_0}\backslash K\right\}.\endaligned
$$
Then $U^{\rho_0}$ admits a fine covering
$$U^{\rho_0}:=\left(\bigcup_{Q_\nu(x,\varepsilon)\in\Cal F_1^{\rho_0}}{Q_\nu(x,\varepsilon)}\right)\cup
\left(\bigcup_{Q_\nu(x,\varepsilon)\in\Cal F_2^{\rho_0}}{Q_\nu(x,\varepsilon)}
\right),$$
and  by Morse's version of Besicovitch's Covering Theorem 
(see \cite{27}, Theorem 5.11)
we may find a subcovering of $U^{\rho_0}$ such that
$$U^{\rho_0}:=\left(\bigcup_{i\in I}Q_i\right)\cup
\left(\bigcup_{j\in I}Q_j\right)\cup N,\qquad K\subset 
\left(\bigcup_{i\in I}\overline{Q_i}\right)\cup N,
$$
 $I$ and $J$ are countable, $Q_i\in \Cal F_1^{\rho_0}$, $Q_j\in \Cal F_2^{\rho_0}$,
the sets $\overline{Q_i}$ and $\overline{Q_j}$ are mutually disjoint, and
$\mu(N)=0$. Then
$$\aligned \eta>\int_U h(x)\,dx\ge \sum_{i\in I}
\int_{Q_i} h(x)\,dx\ge r\sum_{i\in I}\mu(Q_i)
=r\sum_{i\in I}\mu(\overline{Q_i})\ge r \mu(K).
\endaligned$$
By letting $\eta\to 0$ we obtain $\mu(K)=0$, and 
by the inner regularity of $\mu$
we conclude that $\mu(U_0)=0$.
\hfill\qquad\qed
\enddemo
\medskip
\remark{Remark 8.3}
Since for $\mu_s$ a.e. $x_0\in \Omega$
$$\lim_{\varepsilon\to 0^+}\frac{\mu(Q_\nu(x_0,\varepsilon))}
{\varepsilon^N}=\infty,$$
 it is clear that if $x_0\in B_0$ then  $\mu_s$ a.e. $x_0$ is not a Lebesgue point for
$|h|$, otherwise
$$\limsup_{\varepsilon\to 0^+}
 \frac1{\mu(Q_\nu(x_0,\varepsilon))}\int_{Q_\nu(x_0,\varepsilon)}|h(x)| \,dx=
\limsup_{\varepsilon\to 0^+}
 \frac{\varepsilon^N}{\mu(Q_\nu(x_0,\varepsilon))}|h(x_0)|=0. $$
\endremark
\medskip
Using Lemma 8.2, it is possible in some cases to weaken (1.8)
in Proposition 2.1. Indeed, assume that
for $\Cal L^1$ a.e. $u\in\Bbb R$
$$f(\cdot,u,0)\in L^1\loc(\Omega;\Bbb R).\tag8.3$$
Then there exists a countable set $\Cal R_0:=\{r_j\}_j$,
dense in $\Bbb R$,  such that
  $f(\cdot,r_j,0)\in L^1\loc(\Omega;\Bbb R)$ for all $j$.
 Let $\Omega_j$ be the
set of Lebesgue points of $f(\cdot,r_j,0)$ and set
$A_j:=\Omega\backslash \Omega_j$. Then
$$\Cal L^N(\cup_{j=1}^\infty  A_j)=0.$$
Let $B_j$ be the set of points corresponding
 to the set $B_0$ in Lemma 8.2 when
$h:=f(\cdot,r_j,0)$ and $\mu:=|Du|$.
Then
$$|D_s u|(\cup_{j=1}^\infty  B_j)=0.$$
\noindent\proclaim{Lemma 8.4} Proposition 2.1
is still valid provided  we replace $(1.8)$ by $(8.3)$, and
 we take
$$x_0\in \Omega\backslash\left(\bigcup_{i,j=1}^\infty A_j\cup B_i\right),
\qquad t_k:=\cases 1\quad&\text{if $x_0\in \bigcap_{j=1}^\infty\Omega_j$}\\
              |Du|(Q_\nu(x_0,\varepsilon_k))/\varepsilon_k^N
\quad&\text{otherwise},\endcases$$
where $|Du|(\partial Q_\nu(x_0,\varepsilon_k))=0$, $t_k\to T\in(0,\infty]$, 
and 
 $u_0+\alpha_1$, $u_0+\alpha_2\in \Cal R_0$.
\endproclaim
\medskip
\demo{Proof}
The only change is in (2.2). Considering first the case where
$x_0\in \Omega\backslash(\cup_{j=1}^\infty A_j)=\cap_{j=1}^\infty\Omega_j$, 
then $t_k=1$ and (2.2) becomes
$$\aligned 0\le &\int_{E_k^+}
f(x_0+\varepsilon_k y,u_0+\alpha_2,0)\,dy
=\frac1{\varepsilon_k^N}\int_{
Q_\nu(x_0,\varepsilon_k)\cap (x_0+\varepsilon_kE_k^+)}f(x,u_0+\alpha_2,0)\,dx\\
&\le
\frac1{\varepsilon_k^N}\int_{
Q_\nu(x_0,\varepsilon_k)}|f(x,u_0+\alpha_2,0)-f(x_0,u_0+\alpha_2,0)|\,dx
+f(x_0,u_0+\alpha_2,0)\Cal L^N(E_k^+).\endaligned
$$
Since $x_0$ is a Lebesgue point for $f(x,u_0+\alpha_2,0)$ (recall
 that $u_0+\alpha_2\in \Cal R_0$ and that $x_0\in \cap_{j=1}^\infty\Omega_j$)
the first integral on the right hand side approaches zero as $k\to\infty$.
Moreover
$\Cal L^N(E_k^+)\to 0$ as before.\par
If $x_0\in \Omega\backslash(\cup_{i=1}^\infty B_i)$ and 
$x_0\notin \cap_{j=1}^\infty\Omega_j$ then  (2.2) may be estimated as follows:
$$0\le\lim_{k\to\infty}\frac1{t_k}\int_{E_k^+}
f(x_0+\varepsilon_k y,u_0+\alpha_2,0)\,dy\le\lim_{k\to\infty}
\frac1{|Du|(Q(x_0,\varepsilon_k))}\int_{Q(x_0,\varepsilon_k)}
f(x,u_0+\alpha_2,0)\,dx=0,$$
where we have used the fact that $t_k=|Du|(Q_\nu(x_0,\varepsilon_k))
/\varepsilon_k^N$ and the definition of $B_j$ as in (8.3), with 
$h:=f(\cdot,r_j,0)$.
\hfill\qquad\qed
\enddemo
\medskip
We are now ready to state the  main result of this section.

\noindent\proclaim{Theorem 8.5}  Assume that
$f:\Omega\times\Bbb R\times\Bbb R^N\to[0,\infty)$ is a  Borel integrand,
 $f(x,u,\cdot)$  is convex in $\Bbb R^N$, and $f$ satisfies $(8.3)$.
Suppose also that $(1.7)$ holds for all $(x_0,u_0)\in(\Omega\times\Bbb
R)\backslash N_0$, where $N_0$  is a Borel subset of $\Omega\times\Bbb R$. 
Let   $u\in BV\loc(\Omega;\Bbb R)$, and let
$\{u_n\}$ be a sequence of functions in $ W^{1,1}\loc(\Omega;\Bbb R)$
 converging to $u$ in $L^1\loc(\Omega;\Bbb R)$.
\smallskip
\item{\rm (i)} If either  $\Cal H^N(N_0)=0$ or $N_0=M_0\times\Bbb R$
 with $\Cal L^N(M_0)=0$ then
$$\int_\Omega f(x,u(x),\nabla u(x))\,dx
\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx.$$
\item{\rm (ii)} If either  $\Cal H^N(N_0)=0$ or $N_0=M_0\times\Bbb R$ with
$\Cal H^{N-1}(M_0)<\infty$ then
$$\int_\Omega f^\infty(x,u(x),dC(u(x)))\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx.$$
\item{\rm (iii)} If $N_0=M_0\times\Bbb R$ and either
$\Cal H^N(N_0)=0$ or $\Cal H^{N-1}(M_0)=0$ and we assume that
for all $(x_0,u_0)\in(\Omega\times\Bbb R)\backslash N_0$
 and $\varepsilon>0$ there exists
$\delta>0$ such that
$$f(x_0,u,\xi)- f(x,u,\xi)\le\varepsilon(1+f(x,u,\xi))
\tag8.4$$
for  all $(x,u)\in\Omega\times\Bbb R$
 with $|x-x_0|+|u-u_0|\le\delta$ and for all
$\xi\in\Bbb R^ N$, then
$$\int_{S(u)\cap\Omega}\left(
\int_{u^-(x)}^{u^+(x)}f^\infty(x,s,\nu_u)\,ds\right)
d
\Cal H^{N-1}(x)\le\liminf_{n\to\infty}
\int_\Omega f(x,u_n(x),\nabla u_n(x))\,dx.$$
\endproclaim
\medskip
\demo{Proof} (i) We proceed as in Theorem 1.1(i) starting from (2.4).
If $\Cal H^N(N_0)=0$ then in (2.6) we
take $x_0\in\Omega\backslash A$, where $A$ is
the set given in Lemma
8.1(i), otherwise take $x_0\in\Omega\backslash M_0$. Using the notation 
introduced in Lemma 8.1 and thereafter, 
 we may assume, in addition, that $x_0$ is also a Lebesgue point for
all the functions $f(\cdot,r_j,0)$'s, precisely
$$x_0\in \Omega\backslash\left(\cup_{j=1}^\infty A_j\right).$$
 We can now continue with the same argument as in the proof of the Lebesgue part in 
Theorem 1.1,
except that we invoke Lemma 8.4 instead of Proposition 2.1 to justify 
the truncation step.\par (ii) If
$\Cal H^N(N_0)=0$ then take $x_0\in\Omega\backslash B$, where $B$ is the
set given in Lemma 8.1(ii), otherwise take $x_0\in\Omega\backslash M_0$. As
before, let $B_j$ be the
 set introduced in Lemma 8.2 and corresponding to
$f(\cdot,r_j,0)\in L^1\loc(\Omega;\Bbb R)$ (see (8.2)).
Since
$$|C(u)|(\cup_{j=1}^\infty  B_j)=0,$$
 we may assume that  $x_0\in \Omega\backslash (\cup_{j=1}^\infty B_j)$. Now
we continue as in Theorem 1.1(ii),
using  Lemma 8.4 in place of Proposition 2.1, but now
in order to apply Lemma 2.5 of Ambrosio and Dal Maso \cite{4} we first need
 to approximate $g(\xi):= f(x_0,u(x_0),\xi)$
 from below by a nondecreasing sequence of convex functions
which grow at most linearly. This can be done by virtue of 
Proposition 9.1.\par (iii)
Since $N_0=M_0\times\Bbb R$ and $\Cal H^N=\Cal H^{N-1}\times\Cal L^1$ on
$S(u)\times\Bbb R$, it follows that $\Cal H^{N-1}(M_0\cap S(u))=0$. Moreover,
by Lemma 8.2,
$$\Cal H^{N-1}\left(S(u)\cap\left(\cup_{j=1}^\infty B_j\right)\right)=0.$$
Take
$$x_0\in S(u)\backslash \left(M_0\cup\bigcup_{j=1}^\infty B_j\right).$$
We pursue the proof of the jump part  as in Theorem 1.1(iii),
using  Lemma 8.4 instead of Proposition 2.1, but now
in order to apply the density  result of Ambrosio
 we first need
 to approximate $h(u,\xi):= f(x_0,u,\xi)$
 from below by a nondecreasing sequence of  continuous functions
which grow at most linearly. For this purpose we invoke  Proposition 9.3 below.
\hfill\qquad\qed
\enddemo
\medskip\remark{Remark 8.6} (i) The hypothesis placed on part (iii) above, i.e. in the jump part, ensuring that
 the set $N_0$ is of the form $N_0=M_0\times\Bbb R$, is used heavily to apply
 the compactness argument leading to (2.15).\par
(ii) Theorems 1.3, 1.7, 1.8 may be improved similarly to Theorem 8.5 versus 
Theorem 1.1. 
 We leave this to the interested reader.\par
(iii) As in Theorem D of Dal Maso \cite{9}, in the special case where 
$f(x,u,0)\equiv 0$ condition (1.9) can be weakened as follows:\smallskip
{\it
Assume that there exists a set $P_0\subset\Bbb R$ with $\Cal L^1(P_0)=0$ such
that for all $x_0\in \Omega$ and $\varepsilon>0$ there exists
$\delta>0$ such that
$$|f(x_0,u,\xi)- f(x,u,\xi)|\le\varepsilon(1+f(x,u,\xi))
\tag"$(1.9)^\prime$"$$
for  all $x\in\Omega$
 with $|x-x_0|\le\delta$ and for all
$(u,\xi)\in(\Bbb R\backslash P_0)\times\Bbb R^ N$.}
\par The proof of Theorem 1.5 should now be modified
accordingly, using the fact that, if
$A:=\{x\in\Omega\backslash S(u): u(x)\in P_0\}$ then
$\nabla u\equiv 0$ for $\Cal L^N$ a.e. $x\in A$
and $|C(u)|=0$  a.e. $x\in A$. We omit the details.
\endremark



\bigskip

\noindent {\bf\S9. Approximation of convex functions.}
\medskip

Let  $g:\Bbb R^N\to[0,\infty)$ be a convex function. Then 
$$t\mapsto \dfrac{g(t\,\xi)-g(0)}t$$
is increasing, and we define the {\it recession function}
 $$g^\infty(\xi): =\lim_{t\to\infty}\frac{g(t\,\xi)}t=
\sup_{t>0}\frac{g(t\,\xi)-g(0)}t.$$


\noindent\proclaim{Proposition 9.1} 
Let $g:\Bbb R^N\to[0,\infty)$
be a convex function. Then 
$$
\frac{g(t\,\xi)}t\le g^\infty(\xi)+\frac{g(0)}t
\tag9.1
$$
for $t>0$, and there exists an increasing sequence $\{g_j\}_j$ of nonnegative
 convex functions such that:\smallskip
{\rm (i)} $g(\xi)=\sup_{j}g_j(\xi)$ for all $\xi\in \Bbb R^N$;\par
{\rm (ii)} $g^\infty(\xi)=\sup_{j}g_j^\infty(\xi)$ for all $\xi\in \Bbb R^N$;\par
{\rm (iii)} $g_j$ is Lipschitz continuous with Lipschitz constant $j$;\par
{\rm (iv)} if $g(\xi)\ge C(|\xi|-1)$ for some $C>0$, then 
  $g_j$ satisfies the same growth condition for $j \geq[C]+1$.
\endproclaim
\medskip
\demo{Proof} Inequality (9.1) follows immediately from the definition of the $g^{\infty}$.  Define
$$g_j(\xi):=\sup_{|\xi^*|\le j}(\xi^*\cdot\xi-g^*(\xi^*)),$$
 where $g^*$ is the Young--Fenchel conjugate of $g$. Since $g$ is convex, 
$g=g^{**}$; hence $g(\xi)=\sup_{j}g_j(\xi)$. Also, $g_j$ are convex and
$$g_j(\xi)\ge -g^*(0)\ge\inf g\ge 0.$$
This proves (i).

Since $g(\xi)\ge g_j(\xi)$ for each $j$, it follows that 
 $g^\infty(\xi)\ge \sup_j g_j^\infty(\xi)$. Conversely, and by (9.1),
$$\aligned
\frac{g(t\,\xi)}t=\sup_j\frac{g_j(t\,\xi)}t=
\sup_j\left[\frac{g_j(t\,\xi)-g_j(0)}t+\frac{g_j(0)}t\right]\le
\sup_j g^\infty_j(\xi)+\sup_j \frac{g_j(0)}t=
\sup_j g^\infty_j(\xi)+\frac{g(0)}t.\endaligned$$
Letting $t\to\infty$ we conclude that
$g^\infty(t)\le \sup_j g^\infty_j(\xi)$. \par
Property (iii) is straightforward. We prove (iv). If 
$g(\xi)\ge C(|\xi|-1)$ then 
$g^*(\xi^*)\le [C(|\cdot|-1)]^*(\xi^*)$; hence
$$g_j(\xi)=\sup_{|\xi^*|\le j}(\xi^*\cdot\xi-g^*(\xi^*))\ge 
\sup_{|\xi^*|\le j}\left(\xi^*\cdot\xi-[C(|\cdot|-1)]^*(\xi^*)\right).$$
Since
$$[C(|\cdot|-1)]^*(\xi^*)=\cases C\quad&\text{if  $|\xi^*|\le C,$}\\
\infty\quad&\text{otherwise,}
\endcases$$
we conclude that if $j \geq [C]+1$ then
$$
g_j(\xi)\geq  \sup_{|\xi^*|\leq C} (\xi^*\cdot \xi - C) = C(|\xi| - 1).
$$
\hfill\qquad\qed
\enddemo
\medskip

The last proposition of this section uses a corollary of Lindel\"of Theorem which allows us to select a countable collection of functions yielding the supremum function of a noncountable family. For convenience, we include the proof below.

\noindent\proclaim{Lemma 9.2} Let $X$ be a $\sigma$-compact metric space, let $\Cal G\subset C(X;\Bbb R)$, and let $f(x):= \sup_{g \in \Cal G} g(x)$, for all $x \in X$. Then there exists a countable collection $\{g_n\}_{n \in \Bbb N} \subset \Cal G$, such that
$$
f(x) = \sup_n g_n(x) \quad \text{for all}\; x \in X.
$$
\endproclaim  

\demo{Proof} It is clear that $f$ is lower semicontinuous. Therefore, for every $x \in X$ there exist $0 < r(x) < 1$ and $g^1_x \in \Cal G$ such that
$$
f(y) \geq f(x) - \frac12 \; \text{for all}\; y \in B(x, r^1(x)), \quad g^1_x(x) \geq f(x) - \frac12.
$$
Let $\rho^1(x)<r(x)$ be such that
$$
|g^1_x(y) - g^1_x(y')| \leq \frac12 \quad \text{for all}\; y, y' \in B(x,\rho^1(x)).
$$
Since $\{B(x,\rho^1(x))\}_{x\in X}$ is an open covering of $X$, by Lindel\"of Theorem we may extract a countable subcovering $\{B(x_n^1, \rho_n^1)\}$. Recursively, we may find an open covering of $X$,  $\{B(x,\rho^k(x))\}_{x\in X}$, $\rho^k(x)<\frac1{2^k}$, and functions $g^k_x \in \Cal G$, such that for all $x \in X$
$$
f(y) \geq f(x) - \frac1{2^k} \; \text{for all}\; y \in B(x, \rho^k(x)), \; g^k_x(x) \geq f(x) - \frac1{2^k}, \; |g^k_x(y) - g^k_x(y')| \leq \frac1{2^k} \; \text{for all}\; y, y' \in B(x,\rho^k(x)).
$$
Again by Lindel\"of Theorem, $X$ is covered by a countable family $\{B(x_n^k,\rho_n^k)\}$. We claim that
$$
f(x) = \sup_{m,k}\,  g^k_{x_m^k}(x).
$$
Let $x \in X$, $k \in \Bbb N$, and choose  $n \in \Bbb N$  such that $x \in B(x_n^k, \rho_n^k)$. Then
$$
f(x) \geq g^k_{x_n^k}(x) \geq g^k_{x_n^k}(x_n^k) - \frac1{2^k}
\geq f(x_n^k) - \frac2{2^k} \geq \inf_{B\left(x;\frac1{2^k}\right)} f - \frac2{2^k}.
$$
As $f$ is lower semicontinuous, $\liminf_{\varepsilon \to 0} \inf_{B(x,\varepsilon)} \,f = f(x)$, and we conclude that
$$
f(x) \leq \liminf_{k \to \infty}  \inf_{B\left(x;\frac1{2^k}\right)} f - \frac2{2^k} \leq \liminf_{k \to \infty} g^k_{x_n^k}(x) \leq \sup_{m,k}   g^k_{x_m^k}(x) \leq f(x).
$$
\hfill\qquad\qed
\enddemo
\medskip



\noindent\proclaim{Proposition 9.3} {\rm ( \cite{10})}
 Let $A$ be an open set of $\Bbb R^N$ and let
$h:A\times\Bbb R\times\Bbb R^N\to [0,\infty)$ be a function convex in the 
$\xi$ variable,
 and
such that for every  $(x_0,u_0)\in A\times\Bbb R$ and $\varepsilon>0$ there
 exists
$\delta>0$ such that
$$h(x,u,\xi)\ge (1-\varepsilon) h(x_0,u_0,\xi)\tag9.2$$
for  all $(x,u)\in A\times\Bbb R$
 with $|x-x_0|\le\delta$, $|u-u_0|\le\delta$ and for all $\xi\in \Bbb R^ N$.
Then there exists an increasing sequence $\{h_j\}_j$ of nonnegative
 continuous functions, convex in the $\xi$ variable, satisfying (9.2) and such that:\smallskip
{\rm (i)} $h(x,u,\xi)=\sup_{j}h_j(x,u,\xi)$ for all
$(x,u,\xi)\in A\times\Bbb R\times\Bbb R^ N$;\par
{\rm (ii)} $h^\infty(x,u,\xi)=\sup_{j}h_j^\infty(x,u,\xi)$ for all
$(x,u,\xi)\in A\times\Bbb R\times\Bbb R^ N$;\par
{\rm (iii)} $h_j(x,u,\xi)\le C_j(|\xi|+1)$ for all
$(x,u,\xi)\in A\times\Bbb R\times\Bbb R^ N$and for some $C_j>0$;\par
{\rm (iv)} for every  $(x_0,u_0)\in A\times\Bbb R$ and $\varepsilon>0$ there exists
$\delta_j>0$ such that
$$h_j(x,u,\xi)\ge (1-\varepsilon) h_j(x_0,u,\xi),\qquad
h_j(x,u,\xi)\ge (1-\varepsilon) h_j(x,u_0,\xi)$$
for  all $(x,u)\in A\times\Bbb R$
 with $|x-x_0|\le\delta_j$, $|u-u_0|\le\delta_j$, and for all $\xi\in \Bbb R^ N$.
\endproclaim
\medskip
\demo{Proof} Let $\Cal G$ be the class of all continuous
functions
$g:A\times\Bbb R\times\Bbb R^N\to [0,\infty)$,  convex in the $\xi$ variable,
 and
such that:\smallskip\roster
\item $g(x,u,\xi)\le h(x,u,\xi)$ for all
$(x,u,\xi)\in A\times\Bbb R\times\Bbb R^N$;
\item for every  $(x_0,u_0)\in A\times\Bbb R$ and $\varepsilon>0$ there exists
$\delta>0$ such that
$$g(x,u,\xi)\ge (1-\varepsilon) g(x_0,u_0,\xi)$$
and
$$g(x,u,\xi)\ge (1-\varepsilon) g(x_0,u,\xi),\qquad
g(x,u,\xi)\ge (1-\varepsilon) g(x,u_0,\xi)$$
for  all $(x,u)\in A\times\Bbb R$
 with $|x-x_0|\le\delta$, $|u-u_0|\le\delta$, and for all $\xi\in \Bbb R^ N$;
\item there exists $C>0$  such that
$$g(x,u,\xi)\le C(|\xi|+1)\qquad\text{for all
$(x,u,\xi)\in A\times\Bbb R\times\Bbb R^ N$}.$$
\endroster
\par Clearly $\Cal G\ne\emptyset$, as $0\in \Cal G$. 
Following \cite{10},  we claim that
$$h(x_0,u_0,\xi)=\sup_{g\in\Cal G}g(x_0,u_0,\xi)\qquad\text{ for all
$(x_0,u_0,\xi)\in A\times\Bbb R\times\Bbb R^ N$}.\tag9.3$$
By definition of $\Cal G$, it follows immediately that $h\ge \sup_{g\in\Cal G}g$. 
Conversely, fix $(x_0,u_0)\in A\times\Bbb R$,  $\varepsilon>0$, and
 let $\delta$ be such that (9.2) is satisfied. Consider
two cut--off functions $\varphi\in C^\infty_0(A)$,
$\psi\in C^\infty_0(\Bbb R)$, with $0\le\varphi\le 1$, $0\le\psi\le 1$,
 $\varphi\equiv 1$ on $B(x_0,\delta/2)$,
$\varphi\equiv 0$ outside $B(x_0,\delta)$, and, similarly,
$\psi\equiv 1$ on $B(u_0,\delta/2)$,
$\psi\equiv 0$ outside $B(u_0,\delta)$. We can write
$$h(x_0,u_0,\xi)=\sup_jh_j(\xi),$$
where $h_j$ are convex functions satisfying 
the properties stated in Proposition 9.1.
Consider
$$h_j^\varepsilon(x,u,\xi):=(1-\varepsilon)\varphi(x)\psi(u)h_j(\xi).$$
Clearly $h_j^\varepsilon\in\Cal G$ (in particular
property (1) follows from (9.2)). Letting $j\to\infty$ we get
$$(1-\varepsilon)h(x_0,u_0,\xi)=\sup_j h_j^\varepsilon(x_0,u_0,\xi)
\le
\sup_{g\in\Cal G}g(x_0,u_0,\xi);$$
hence the claim follows by letting $\varepsilon\to 0^+$.\par
By Lemma 9.2 and (9.3)
 there exist a sequence $h_j$ in $\Cal G$ such that
$h(x,u,\xi)=\sup_{j}h_j(x,u,\xi)$ for all
$(x,u,\xi)$ in $A\times\Bbb R\times\Bbb R^ N$.
 Due to the stability properties of the class $\Cal G$, we can assume
the sequence $\{h_j\}_j$ increasing.
 Indeed it is easy to see that if $g_1$, $g_2\in\Cal G$ then
$g_1\vee g_2\in\Cal G$ (while in general
$g_1\wedge g_2\notin\Cal G$, since we may loose convexity). This proves (i). Clearly, properties (iii) and (iv) follow immediately from the definition of $\Cal G$.    
\par
(ii) follows easily from Proposition 9.1, (9.1), and the fact that clearly
$h^\infty(x,u,\xi)\ge\sup_{j}h_j^\infty (x,u,\xi)$. Indeed,
$$\frac{h(x,u,t\,\xi)}t=\sup_j\left[\frac{h_j(x,u,t\,\xi)-h_j(x,u,0)}t+
\frac{h_j(x,u,0)}t\right]\le \sup_j h_j^\infty(x,u,\xi)+
\frac{h(x,u,0)}t,$$
so letting $t\to\infty$ we obtain (ii). 
\hfill\qquad\qed
\enddemo



\bigskip


\noindent {\bf Acknowledgments.} The research of I. Fonseca was
partially supported by the National Science Foundation through the
Center for Nonlinear Analysis under Grants No. DMS--9500531 and
DMS--9731957, and that of both authors by the Max--Planck Institute
for Mathematics in the Sciences, Leipzig (Germany).
\bigskip

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\enddocument
