#### Proof.

Consider $0<\epsilon <{\epsilon}_{0}\le 1$ as in hypothesis (4), then $u\in {W}^{1,p}({B}_{R+2{\epsilon}_{0}};{\mathbb{R}}^{N})$ and

$\mathcal{\mathcal{F}}(u,{B}_{R+2{\epsilon}_{0}})={\int}_{{B}_{R+2{\epsilon}_{0}}}f(x,Du(x))\mathit{d}x<+\mathrm{\infty}.$

Let us denote ${u}_{\epsilon}(x):=(u*{\varphi}_{\epsilon})(x)$, the usual mollification, where $x\in {B}_{R}$, and define

${f}_{\epsilon}(x,z)=\underset{y\in \overline{B(x,\epsilon )}}{\mathrm{min}}f(y,z).$(3.2)

By definition, it follows that

$|D{u}_{\epsilon}(x)|\le ({\int}_{{B}_{R+2{\epsilon}_{0}}}{|Du(y){|}^{p}dy)}^{\frac{1}{p}}({\int}_{{\mathbb{R}}^{n}}{|{\varphi}_{\epsilon}(y){|}^{{p}^{\prime}})}^{\frac{1}{{p}^{\prime}}}\le C{\epsilon}^{-\frac{n}{p}},$

where $C=C({\parallel Du\parallel}_{{L}^{p}})>1$.
Moreover, by the Hölder continuity hypothesis (i.e., hypothesis 2), we have

${f}_{\epsilon}(x,z)\ge f(x,z)-H{\epsilon}^{\alpha}(1+{|z|}^{q}).$(3.3)

Note that the left-hand side of hypothesis (1) gives

${|z|}^{p}\le {f}_{\epsilon}(x,z).$(3.4)

Now we observe that is possible to find $K=K(p,q,{\parallel Du\parallel}_{{L}^{p}},H)<+\mathrm{\infty}$ such that

$f(x,z)\le K{f}_{\epsilon}(x,z)+H,x\in {B}_{R},|z|\le C{\epsilon}^{-\frac{n}{p}}.$(3.5)

Indeed, let us fix $\delta \in (0,1)$ and observe that, using (3.3), (3.4) and $|z|\le C{\epsilon}^{-\frac{n}{p}}$, we get

${f}_{\epsilon}(x,z)=\delta {f}_{\epsilon}(x,z)+(1-\delta ){f}_{\epsilon}(x,z)$$\ge \delta f(x,z)-\delta H{\epsilon}^{\alpha}(1+{|z|}^{q})+(1-\delta ){|z|}^{p}$$=\delta f(x,z)-\delta H{\epsilon}^{\alpha}{|z|}^{q}+(1-\delta ){|z|}^{p}-\delta H{\epsilon}^{\alpha}$$=\delta f(x,z)-\delta H{\epsilon}^{\alpha}{|z|}^{p}{|z|}^{q-p}+(1-\delta ){|z|}^{p}-\delta H{\epsilon}^{\alpha}$$\ge \delta f(x,z)-\delta {C}^{q-p}H{\epsilon}^{\alpha +(\frac{p-q}{p})n}{|z|}^{p}+(1-\delta ){|z|}^{p}-\delta H{\epsilon}^{\alpha}$$\ge \delta f(x,z)+(1-\delta -\delta {C}^{q-p}H){|z|}^{p}-\delta H,$

where the last estimate relies on the fact that $\frac{q}{p}\le \frac{n+\alpha}{n}$, $0<\alpha \le 1$ and $0<\epsilon <1$.
Then (3.5) follows choosing $K=\frac{1}{\delta}=1+{C}^{q-p}H$. Now, using hypothesis (4), Jensen’s inequality and (3.2), we obtain

${f}_{\epsilon}(x,D{u}_{\epsilon}(x))=f(\stackrel{~}{y},D{u}_{\epsilon}(x))$$\le {\displaystyle {\int}_{B(x,\epsilon )}}f(\stackrel{~}{y},Du(y)){\varphi}_{\epsilon}(x-y)\mathit{d}y$$\le {\displaystyle {\int}_{B(x,\epsilon )}}f(y,Du(y)){\varphi}_{\epsilon}(x-y)\mathit{d}y$$=\left(f(\cdot ,Du(\cdot ))*{\varphi}_{\epsilon}\right)(x)$$=:f{(\cdot ,Du(\cdot ))}_{\epsilon}(x).$(3.6)

Therefore, using (3.5), we have

$f(x,D{u}_{\epsilon}(x))\le Kf{(\cdot ,Du(\cdot ))}_{\epsilon}(x)+H.$

Finally, since $f{(\cdot ,Du(\cdot ))}_{\epsilon}(x)\to f(x,Du(x))$ strongly in ${L}^{1}({B}_{R})$, by recalling that ${u}_{\epsilon}\to u$ in ${W}^{1,p}({B}_{R};{\mathbb{R}}^{N})$, and by using a well-known variant of Lebesgue’s dominated convergence theorem and Lemma 2.2, the proof is completed.
∎

#### Proof.

We follow the proof of [9, Theorem 6].
By the proof of Theorem 3.1, if $u\in {W}_{\mathrm{loc}}^{1,p}(\mathrm{\Omega};{\mathbb{R}}^{N})$ is such that $h(x,Du(x))\in {L}_{\mathrm{loc}}^{1}(\mathrm{\Omega})$, then there exists a sequence ${\{{u}_{m}\}}_{m\in \mathbb{N}}\subset {W}^{1,q}({B}_{R};{\mathbb{R}}^{N})$ such that ${u}_{m}\to u$ strongly in ${W}^{1,p}({B}_{R};{\mathbb{R}}^{N})$, $D{u}_{m}(x)\to Du(x)$ a.e., $h(x,D{u}_{m}(x))\to h(x,Du(x))$ a.e., and ${\int}_{{B}_{R}}h(x,D{u}_{m}(x))\to {\int}_{{B}_{R}}h(x,Du(x))$.
Using a well-known variant of Lebesgue’s dominated convergence theorem, we have that

${\int}_{{B}_{R}}f(x,D{u}_{m}(x))\to {\int}_{{B}_{R}}f(x,Du(x)),$

and then, by Lemma 2.2, $\mathcal{\mathcal{L}}(u,{B}_{R})=0$.
∎

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