Proof. By the assumption $\mu$ can be represented with the use of $f\in L^{1}(\Omega )$ and $G\in (L^{\widetilde \varphi (\cdot )}(\Omega ))^{n}$, such that $\mu =f-{{\rm div\,}}G$ in the sense of distributions. Let us fix arbitrary (small) ${{\varepsilon }}>0$. We consider a sequence of truncations $\{T_\ell f\}_\ell$. Note that $\{T_\ell f\}_\ell$ converges to $f$ strongly in $L^{1}(\Omega )$, so we can choose $k$ large enough for $\|T_kf-f\|_{L^{1}(\Omega )}<{{\varepsilon }}/2$.

Note that for any $\widetilde {{\varepsilon }}>0$ there exists an open set $A\supset E$ with ${{{\rm cap}_{\varphi (\cdot )}}}(A,\Omega )<\widetilde {{\varepsilon }}$. Parameter $\widetilde {{\varepsilon }}$ will be chosen in a few lines. Let us fix a compact set $K\subset A$. By definition of ${{{\rm cap}_{\varphi (\cdot )}}}$ there exists a sequence $\{\phi _j\}_j\subset C_0^{\infty }(A)$ of functions such that

\[ K\subset\{\phi_j= 1\},\quad 0\leq\phi_j\leq 1 \text{ and for }j\geq j_0\ \| \nabla\phi_j\|_{(L^{\varphi({\cdot})^{n}})(A)}\leq 2\widetilde{{\varepsilon}}. \]

Indeed to restrict to $\phi _j$ such that $0\leq \phi _j\leq 1$ let us point out that the map $t\mapsto {\rm min}\{t,1\}$ is Lipschitz, so $\widetilde f := {\rm min}\{ f,1\}\in \mathcal {R}_{\varphi (\cdot )}(K,\Omega )$ and ${{\varrho }}_{\varphi (\cdot ),K}(|\nabla \widetilde f|)\leq {{\varrho }}_{\varphi (\cdot ),K}(|\nabla f|)$. On the other hand, we substituted the modular used in definition of ${{{\rm cap}_{\varphi (\cdot )}}}$ with the norm. This is justified by the fact that the sequence realizing the infimum converges also in norm topology, which follows from the doubling growth of $\varphi$ via (2.5).

We note that

\[ |\mu|(K)\leq \left|\int_{A}\phi_j\,{\rm d}\mu\right|=\left|\int_{A}f\,\phi_j\,{\rm d}x + \int_{A}G\cdot\nabla \phi_j\,{\rm d}x\right|. \]

Due to Hölder inequality (2.4) and then the Poincaré inequality (2.6), we have

\begin{align*} |\mu|(K)& \leq \int_{A}|T_kf-f|\,|\phi_j|\,{\rm d}x+\int_{A}|T_kf|\,|\phi_j|\,{\rm d}x + \int_{A}|G\cdot\nabla \phi_j|\,{\rm d}x\\ & \leq \|T_kf-f\|_{L^{1}(A)}\,\|\phi_j\|_{L^{\infty}(A)}+2\|T_kf\|_{L^{\widetilde\varphi({\cdot})}(A)}\,\|\phi_j\|_{L^{\varphi({\cdot})}(A)} \\ & \quad + 2\|G\|_{(L^{\widetilde\varphi({\cdot})}(A))^{n}}\,\|\nabla \phi_j\|_{(L^{\varphi({\cdot})}(A))^{n}} \\ & \leq \|T_kf-f\|_{L^{1}(A)} +2\left(\|T_kf\|_{L^{\widetilde\varphi({\cdot})}(A)} + \|G\|_{(L^{\widetilde\varphi({\cdot})}(\Omega))^{n}}\right)\|\nabla \phi_j\|_{(L^{\varphi({\cdot})}(A))^{n}} . \end{align*}

We pick $\widetilde {{\varepsilon }}={{\varepsilon }}/(8(2k + \|G\|_{(L^{\widetilde \varphi (\cdot )}(\Omega ))^{n}}))$. Then $|\mu |(K)<{{\varepsilon }}$ with arbitrary ${{\varepsilon }}>0$, so $|\mu |(K)=0$. Therefore

\[ |\mu|(E)\leq|\mu|(A)=\sup\{|\mu|(K):\ K\subset A,\ K \text{ compact}\}=0, \]

which ends the proof.

Proof of theorem 1.1. Lemma 3.1 provides the implication: if $\mu$ belongs to $( L^{1}(\Omega )+(W_0^{1,{\varphi (\cdot )}}(\Omega ))')\cap {{\mathcal {M}_b}}(\Omega )$, then $\mu \in \mathcal {M}^{\varphi (\cdot )}_b(\Omega )$. Therefore, we concentrate now on the essentially more demanding converse, that is, if $\mu _{\varphi (\cdot )}\in \mathcal {M}^{\varphi (\cdot )}_b(\Omega )$, then $\mu _{\varphi (\cdot )}\in L^{1}(\Omega ) + (W^{1,{\varphi (\cdot )}}_0(\Omega ))'$.

Step 1. Initial decomposition. We will show that for a nonnegative ${{\mu _{\varphi (\cdot )}}}\in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$ there exists a nonnegative measure ${{\gamma ^{\rm meas}}}\in (W_0^{1,{\varphi (\cdot )}}(\Omega ))'$ and nonnegative Borel measurable function $h\in L^{1}(\Omega,{{\gamma ^{\rm meas}}})$ such that ${\rm d}{{\mu _{\varphi (\cdot )}}}=h\,{\rm d}{{\gamma ^{\rm meas}}}$.

For any $\widetilde u\in W^{1,{\varphi (\cdot )}}(\Omega )$ we can find its ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-continuous representative denoted by $u$ (lemma 2.7). We define a functional $\mathcal {F}: W^{1,{\varphi (\cdot )}}_0(\Omega )\to [0,\infty ]$ by

\[ \mathcal{F}[u]=\int_\Omega u_+\,{\rm d}{{\mu_{\varphi({\cdot})}}} \]

and observe that it is convex and lower semicontinuous on a separable space $W^{1,{\varphi (\cdot )}}_0(\Omega )$. Thus, $\mathcal {F}$ can be expressed as a supremum of a countable family of continuous affine functions [Reference Chlebicka, Gwiazda, Wróblewska-Kamińska and Świerczewska-Gwiazda26, lemma 2.1.11]. By its very definition $(W^{1,{\varphi (\cdot )}}_0(\Omega ))'$ consists of all linear functionals on $W^{1,{\varphi (\cdot )}}_0(\Omega )$. Therefore, there exist sequences of functions $\{\xi _n\}_n\subset (W^{1,{\varphi (\cdot )}}_0(\Omega ))'$ and numbers $\{a_n\}_n\subset {{\mathbb {R}^{n}}}$ such that

\[ \mathcal{F}[u]=\sup_{n\in\mathbb{N}}\,\{\langle\xi_n,u\rangle-a_n\}\quad\text{for all }\ u\in W^{1,{\varphi({\cdot})}}_0(\Omega). \]

Then, for any $s>0$, $s\mathcal {F}[u]=\mathcal {F}[su]\geq s\langle \xi _n,u\rangle -a_n$ for every $n$. When we divide it by $s$ and let $s\to \infty$ we obtain that $\mathcal {F}[u]\geq \langle \xi _n,u\rangle$ for all $u\in W_0^{1,{\varphi (\cdot )}}(\Omega )$. As $\mathcal {F}[0]= 0$ we infer that $a_n\geq 0$. Therefore, $\mathcal {F}[u]\geq \sup _{n\in \mathbb {N}} \langle \xi _n,u\rangle \geq \sup _{n\in \mathbb {N}}\{\langle \xi _n,u\rangle -a_n\}=\mathcal {F}[u]$ and, in turn,

(3.1)\begin{equation} \mathcal{F}[u]=\sup_{n\in\mathbb{N}}\, \langle\xi_n,u\rangle. \end{equation}

This means that for all $\phi \in C_0^{\infty }(\Omega )$ we have

\[ \langle\xi_n,\phi\rangle\leq \sup_{n\in\mathbb{N}}\,\langle\xi_n,\phi\rangle=\mathcal{F}[\phi]=\int_\Omega \phi_+\,{\rm d}{{\mu_{\varphi({\cdot})}}}\leq \|{{\mu_{\varphi({\cdot})}}}\|_{{{\mathcal{M}_b}}(\Omega)} \|\phi\|_{L^{\infty}(\Omega)}. \]

By the same arguments for $-\varphi$ we get

\[ |\langle\xi_n,\phi\rangle|\leq \|{{\mu_{\varphi({\cdot})}}}\|_{{{\mathcal{M}_b}}(\Omega)} \|\phi\|_{L^{\infty}(\Omega)} \]

implying that $\xi _n\in (W^{1,{\varphi (\cdot )}}_0 (\Omega ))'\cap {{\mathcal {M}_b}}(\Omega ).$ By the Riesz representation theorem there exists nonnegative ${{\xi _n^{\rm meas}}}\in {{\mathcal {M}_b}}(\Omega )$, such that

\[ \langle\xi_n,\phi\rangle=\int_\Omega \phi \ {\rm d}{{\xi_n^{\rm meas}}} \quad\text{for all }\phi\in C_0^{\infty}(\Omega). \]

Note that

(3.2)\begin{equation} {{\xi_n^{\rm meas}}}\leq{{\mu_{\varphi({\cdot})}}}\quad\text{and}\quad\|{{\xi_n^{\rm meas}}}\|_{{{\mathcal{M}_b}}(\Omega)}\leq \|{{\mu_{\varphi({\cdot})}}}\|_{{{\mathcal{M}_b}}(\Omega)}. \end{equation}

Let us define

(3.3)\begin{equation} \eta=\sum_{n=1}^{\infty}\frac{\xi_n}{2^{n}(\|\xi_n\|_{(W^{1,{\varphi({\cdot})}}_0(\Omega))'}+1)} \end{equation}

and observe that the series in absolutely convergent in $(W^{1,{\varphi (\cdot )}}_0(\Omega ))'$. Therefore, for $\phi \in C_0^{\infty }(\Omega )$ we can write

\begin{align*} |\langle\eta,\phi\rangle|& \leq \sum_{n=1}^{\infty}\frac{|\langle \xi_n,\phi\rangle|}{2^{n}(\|\xi_n\|_{(W^{1,{\varphi({\cdot})}}_0(\Omega))'}+1)}\\ & \leq \sum_{n=1}^{\infty}\frac{\|{{\xi_n^{\rm meas}}}\|_{{{\mathcal{M}_b}}(\Omega)}}{2^{n}}\|\phi\|_{L^{\infty}(\Omega)} \leq \|{{\mu_{\varphi({\cdot})}}}\|_{{{\mathcal{M}_b}}(\Omega)}\|\phi\|_{L^{\infty}(\Omega)} \end{align*}

and so $\eta \in (W^{1,{\varphi (\cdot )}}_0(\Omega ))'\cap {{\mathcal {M}_b}}(\Omega )$ too. We denote

\[ {{\eta^{\rm meas}}}=\sum_{n=1}^{\infty}\frac{{{\xi_n^{\rm meas}}}}{2^{n}(\|\xi_n\|_{(W^{1,\varphi({\cdot})}_0(\Omega))'}+1)} \]

and note that this is a series of positive elements that is absolutely convergent in ${{\mathcal {M}_b}}(\Omega )$. Moreover, ${{\xi _n^{\rm meas}}}\ll {{\eta ^{\rm meas}}}$ and thus for every $n$ there exists a nonnegative function $h_n\in L^{1}(\Omega,{\rm d}{{\eta ^{\rm meas}}})$ such that ${\rm d}{{\xi _n^{\rm meas}}}=h_n\, {\rm d}{{\eta ^{\rm meas}}}$. Having (3.1) we deduce that

(3.4)\begin{align} \langle{{\mu_{\varphi({\cdot})}}},\phi\rangle& =\int_\Omega \phi\,{\rm d}{{\mu_{\varphi({\cdot})}}}=\sup_{n\in\mathbb{N}}\int_\Omega \phi\,{\rm d}{{\xi_n^{\rm meas}}}\notag\\ & =\sup_{n\in\mathbb{N}}\int_\Omega h_n\,\phi\,{\rm d}{{\eta^{\rm meas}}}\qquad\text{for any }\ \phi\in C_0^{\infty}(\Omega). \end{align}

On the other hand, (3.2) ensures that $h_n{{\eta ^{\rm meas}}}\leq {{\mu _{\varphi (\cdot )}}}$. This means that for any measurable set $E\subset \Omega$ and every $n$ we have

(3.5)\begin{equation} \int_E h_n\,{\rm d} {{\eta^{\rm meas}}}\leq {{\mu_{\varphi({\cdot})}}}(E).\end{equation}

We denote $h_{\max }^{k}=\max \{h_1(x),\dots,h_k(x)\}$ and

(3.6)\begin{equation} E^{j,k}=\{x\in E:\ \ h_{\max}^{j}(x)>h_i(x)\ \text{ for every }\ i=1,\dots,k-1\}. \end{equation}

Then $E^{j,k}$ for $i=1,\dots,k$ are pairwise disjoint and $E=\cup _{j=1}^{k} E^{j,k}$, so

(3.7)\begin{equation} \int_E h_{\max}^{k}(x)\,{\rm d} {{\eta^{\rm meas}}}\leq \sum_{j=1}^{k}\int_{E^{j,k}} h_{\max}^{k}(x)\,{\rm d} {{\eta^{\rm meas}}}\leq \sum_{j=1}^{k} {{\mu_{\varphi({\cdot})}}}(E^{j,k})={{\mu_{\varphi({\cdot})}}}(E).\end{equation}

Let us pass with $k\to \infty$ and take $h(x)=\sup _{l\in \mathbb {N}} h_l(x)$. We infer that for any measurable set $E\subset \Omega$

(3.8)\begin{equation} \int_E h\,{\rm d} {{\eta^{\rm meas}}}\leq {{\mu_{\varphi({\cdot})}}}(E).\end{equation}

According to (3.4), for every nonnegative $\phi \in C_0^{\infty }$ we have

\[ \int_\Omega \phi\,{\rm d}{{\mu_{\varphi({\cdot})}}}=\sup_{l\in\mathbb{N}} \int_\Omega h_l\, \phi\,{\rm d}{{\eta^{\rm meas}}}\leq \int_\Omega h \,\phi\,{\rm d}{{\eta^{\rm meas}}}\leq \int_\Omega \phi\,{\rm d}{{\mu_{\varphi({\cdot})}}} \]

that is ${\rm d}{{\mu _{\varphi (\cdot )}}}=h\,{\rm d}{{\eta ^{\rm meas}}}.$ Since ${{\mu _{\varphi (\cdot )}}}(\Omega )\in {{\mathcal {M}_b}}(\Omega )$, we deduce that $h\in L^{1}(\Omega,{\rm d}{{\eta ^{\rm meas}}})$ and the aim of this step is achieved with ${{\gamma ^{\rm meas}}}={{\eta ^{\rm meas}}}\in (W^{1,{\varphi (\cdot )}}_0(\Omega ))'$.

Step 2. Auxiliary sequence of measures. We take an increasing sequence of sets $\{K_i\}_i$ compact in $\Omega$, such that $\cup _{i=1}^{\infty } K_i=\Omega$. We set

\[ \widetilde\mu_i=T_i(h\mathbb {1}_{K_i}){{\gamma^{\rm meas}}}\quad\text{for every }\ i\in\mathbb{N}. \]

Then $\{\widetilde \mu _i\}_i$ is an increasing sequence of positive measures in $(W^{1,{\varphi (\cdot )}}_0(\Omega ))'$ with supports compact in $\Omega$. Let us denote

\[ \mu_0 =\widetilde\mu_0\quad\text{ and }\quad \mu_i=\widetilde\mu_i-\widetilde\mu_{i-1}\quad\text{for every }\ i\in\mathbb{N}. \]

Then $\sum _{i=1}^{k}\mu _i=T_k(h\mathbb {1}_{K_k}){{\gamma ^{\rm meas}}}\in {{\mathcal {M}_b}}(\Omega )$. Since $\mu _i\geq 0$, we have $\sum _{i=1}^{\infty } \|\mu _i\|_{\mathcal {M}(\Omega )}<\infty$. Furthermore, ${{\mu _{\varphi (\cdot )}}}=\sum _{i=1}^{\infty }\mu _i$ and this series is absolutely convergent in ${{\mathcal {M}_b}}(\Omega )$.

Step 3. Construction of decomposition. Suppose ${{\varrho }}\in C_0^{\infty }(B(0,1))$ is a standard mollifier (nonnegative and symmetric function with $\int _{{\mathbb {R}^{n}}} {{\varrho }}\,{\rm d}x=1$) and set ${{\varrho }}_k(x)=k^{n}{{\varrho }}(kx)$. We consider mollification

\[ \mu_{i,k}^{{\varrho}}(x)=\int_{{\mathbb{R}^{n}}} {{\varrho}}_k(x-y)\,{\rm d}\mu_i(y ) \]

For $k$ large enough, we can decompose $\mu _i=f_i+w_i$ with

\[ f_i=\mu_{i,k_i}^{{\varrho}}\in C_0^{\infty}(\Omega)\quad\text{and}\quad w_i=\mu_i-\mu_{i,k_i}^{{\varrho}}\in (W^{1,{\varphi({\cdot})}}_0(\Omega))' \]

by choosing for every $i$ sufficiently large $k_0^{i}$ such that for $k_i>k_0^{i}$, $\mu _{i,k_i}^{{\varrho }}$ belongs to $C_0^{\infty }(\Omega )$, so we restrict our attention to such $k_i$. Therefore, we get – up to a subsequence – convergence of $\{f_{i}\}_i=\{\mu _{i,k_i}^{{\varrho }}\}_i$ to ${{\mu _{\varphi (\cdot )}}}$ in measure and for every $i$ we have $\|f_i\|_{L^{1} (\Omega )}\leq \|{{\mu _{\varphi (\cdot )}}}\|_{{{\mathcal {M}_b}}(\Omega )}.$ By step 2 the series $\sum _{i=1}^{\infty } f_i$ is convergent in $L^{1} (\Omega )$ and there exists its limit $f^{0}=\sum _{i=1}^{\infty } f_i\in L^{1}(\Omega )$. As for convergence of $w_i$ we observe first that due to [Reference Harjulehto and Hästö49, lemma 6.4.5] we get convergence of $\{\mu _{i,k}^{{\varrho }}\}_k$ to $\mu _i$ in $(W^{1,{\varphi (\cdot )}}_0(\Omega ))'$ as $k\to \infty$. We note that the series $\sum _{i=1}^{\infty } w_i$ converges in $(W^{1,{\varphi (\cdot )}}_0(\Omega ))'$ and there exists its limit $w^{0}=\sum _{i=1}^{\infty } w_i\in (W^{1,{\varphi (\cdot )}}_0(\Omega ))'$. Therefore, the three following series converge in the sense of distributions

\[ \sum_{i=1}^{\infty} \mu_i={{\mu_{\varphi({\cdot})}}},\quad \sum_{i=1}^{\infty} f_i=f^{0}\quad \text{and} \quad\sum_{i=1}^{\infty} w_i=w^{0} \]

and, consequently, ${{\mu _{\varphi (\cdot )}}}=f^{0}+w^{0}$.

Step 4. Summary. Let us recall that the proof starts with justification that for a nonnegative measure $\mu \in L^{1}(\Omega )+(W^{1,{\varphi (\cdot )}}_0(\Omega ))'\subset {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega ) .$ Step 3 provides the reverse implication.

If the measure was not nonnegative, we can write that ${{\mu _{\varphi (\cdot )}}}=(({{\mu _{\varphi (\cdot )}}})_++({{\mu _{\varphi (\cdot )}}})_-)$ and repeat the above reasoning separately for its positive and negative part. Note that by the monotonicity of capacity, if ${{{\rm cap}_{\varphi (\cdot )}}}(A,\Omega )=0$, then $({{\mu _{\varphi (\cdot )}}})_+(A)=0=({{\mu _{\varphi (\cdot )}}})_-(A)$ ($\varphi (\cdot$)-capacity can be exhausted over Borel sets included in $A$), see [Reference Baruah, Harjulehto and Hästo10]. Note that if $\mu _{\varphi (\cdot )}$ is nonnegative, then $f\geq 0$. Hence, for a signed measure $\mu \in {{\mathcal {M}_b}}(\Omega )$ we infer that $\mu \in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$ if and only if $\mu \in ( L^{1}(\Omega )+(W^{1,{\varphi (\cdot )}}_0(\Omega ))')\cap {{\mathcal {M}_b}}(\Omega )$, that is when there exists $f\in L^{1}(\Omega )$ and $G\in (L^{\widetilde \varphi (\cdot )}(\Omega ))^{n}$, such that ${{\mu _{\varphi (\cdot )}}}=f-{{\rm div\,}}G\ \text { in the sense of distributions}.$ Hence, the proof of the capacitary characterization is completed.

Proof. We fix an arbitrary sequence $D_1\subset D_2\subset \dots \subset \mathfrak {M}$ of sets with ${{{\rm C}_{\varphi (\cdot )}}}(D_i)=0$, such that

\[ \lim_{i\to\infty}\mu(D_i)=\alpha:=\sup\{\mu(D):\ D\in\mathfrak{M}\ \text{ and }\ {{{\rm C}_{\varphi({\cdot})}}}(D)=0\}<\infty. \]

Denote $D_\infty =\bigcup _{i=1}^{\infty } D_i$ and note that $D_\infty \in \mathfrak {M}$, ${{{\rm C}_{\varphi (\cdot )}}}(D_\infty )=0$ and $\mu (D_\infty )=\alpha$. Let us observe that $\mu (D\setminus D_\infty )=0$ for every $D\in \mathfrak {M}$ with ${{{\rm C}_{\varphi (\cdot )}}}(D)=0$. By setting

\[ \mu_{ac}=\mathbb {1}_{{{\mathbb{R}^{n}}}\setminus D_\infty}\mu\quad\text{and}\quad {{\mu_{\rm sing}}}=\mathbb {1}_{D_\infty}\mu \]

we get the decomposition of the desired properties. Due to [Reference Baruah, Harjulehto and Hästo10, proposition 7] countable sum of sets of $W^{1,\varphi (\cdot )}$-capacity zero is of $W^{1,\varphi (\cdot )}$-capacity zero, so the uniqueness of the decomposition ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-everywhere is evident.

Proof of proposition 4.2. To get (4.2), we use first ($\mathcal {A}$2)$_1$, (4.1) tested by $T_k(u_s)\in W^{1,\varphi (\cdot )}_0(\Omega )$, and the above remark, in the following way

\begin{align*} c_1^{\varphi}\int_\Omega\varphi(x,|\nabla T_k(u_s)|)\,{\rm d}x& \leq \int_\Omega \mathcal{A}(x,\nabla T_k(u_s))\cdot \nabla T_k(u_s)\,{\rm d}x\\ & = \int_\Omega \mathcal{A}(x,\nabla u_s)\cdot \nabla T_k(u_s)\,{\rm d}x\\ & =\int_\Omega T_k(u_s)\,{\rm d}\mu^{s} \leq 2 k\|\mu^{s}\|_{\mathcal{M}_b(\Omega)}. \end{align*}

We conclude the last inequality above because of the assumed properties of $\mu ^{s}$.

In order to get (4.3), we use ($\mathcal {A}$2)$_2$, doubling growth (2.3), and finally (4.2) to conclude that for any $k>1$ we have

(4.5)\begin{align} & \int_\Omega\widetilde\varphi\big(x,|\mathcal{A}(x,\nabla T_k(u_s))|\big)\,{\rm d}x\leq \int_\Omega \widetilde\varphi\Big(x,1+\gamma(x)+\varphi(x,|\nabla T_k(u_s)|)/|\nabla T_k(u_s)|\Big)\,{\rm d}x\nonumber\\ & \quad\leq \frac{1}{2} \left( \int_\Omega \widetilde\varphi\Big(x,2\big(1+\gamma(x)\big)\Big)\,{\rm d}x+\int_\Omega \widetilde\varphi\Big(x,\varphi\big(x,2|\nabla T_k(u_s)|\big)/|\nabla T_k(u_s)|\Big)\,{\rm d}x\right)\nonumber\\ & \quad\leq c\left(\int_\Omega \widetilde\varphi\big(x,1+\gamma(x)\big)\,{\rm d}x+\int_\Omega \varphi\big(x, |\nabla T_k(u_s)|\big)\,{\rm d}x\right)\leq c\,k, \end{align}

where $c= c(c_1^{\varphi },c_2^{\varphi },c_{\Delta _2}(\varphi ),q,\|\mu \|_{\mathcal {M}_b(\Omega )}, \|1+\gamma \|_{L^{\widetilde \varphi (\cdot )}(\Omega )})$.

To get (4.4) we start with observing that $|\{|u_s|\geq k\}|=|\{T_k(|u_s|)\geq k\}|$. Then by Tchebyshev inequality, Poincaré inequality and (4.2) as follows

\begin{align*} |\{|u_s|\geq k\}|& \leq \int_\Omega\frac{|T_k(u_s)|^{p}}{k^{p}}{\rm d}x\leq \frac{c}{k^{p}}\int_\Omega |\nabla T_k(u_s)|^{p}\,{\rm d}x\\ & \leq \frac{c}{k^{p}}\int_\Omega \varphi(x,|\nabla T_k(u_s)|)\,{\rm d}x \leq c k^{1-p}\xrightarrow[k\to\infty]{}0. \end{align*}

Proof. Having (4.2) we get that $\{T_k(u_s)\}_s$ is uniformly bounded in $W^{1,\varphi (\cdot )}_{0}(\Omega )$. By recalling the Banach–Alaoglu theorem in the reflexive space, we infer that there exists a (non-relabelled) subsequence of $\{u_s\}$ and function $u\in \mathcal {T}^{1,\varphi (\cdot )}_0$ such that for $s\to 0$ we have (5.3). Note that in the general case we would have here weak-$\ast$ convergence, but our space is reflexive and these notions of convergences coincide. Since embedding (2.7) is compact, up to a non-relabelled subsequence, for $s\to 0$ also (5.2). Consequently, due to the Dunford–Pettis theorem, up to an (again) non-relabelled subsequence $\{T_k(u_s)\}_s$ is a Cauchy sequence in measure and (5.1) holds. By the same arguments, due to (4.3), there exists $\mathcal {A}_k^{\infty }\in (L^{\widetilde \varphi (\cdot )}(\Omega ))^{n}$ such that for $s\to 0$

(5.8)\begin{equation} \mathcal{A}(x,\nabla T_k(u_s))\rightharpoonup{}\mathcal{A}_k^{\infty} \quad\text{weakly in }\ (L^{\widetilde{\varphi}({\cdot})}(\Omega))^{n}\quad \text{for every }k>0. \end{equation}

The effort will be put now in identification of the limit function

(5.9)\begin{equation} \mathcal{A}_k^{\infty} = \mathcal{A}(x,\nabla T_k(u))\quad \text{a.e. in }\Omega,\text{ for every }k>0 \end{equation}

and proving that $u$ obtained in this procedure is a very weak solution. Recall that $\mathcal {A}$ is continuous with respect to the last variable, we have the convergence (5.2) and what remains to prove is fine behaviour of $\{\nabla u_s\}_s$. In order to show that $\{\nabla u_s\}_s$ is a Cauchy sequence in measure we set ${{\epsilon }} >0$ and $m,n\in \mathbb {N}$ arbitrary (large). Given any $t, \tau, r >0$, one has that

(5.10)\begin{align} |\{ |\nabla u_l-\nabla u_m|>t\}| & \leq |\{|\nabla u_l|>\tau\}|+|\{|\nabla u_m|>\tau\}|+|\{|u_l|>\tau\}|\nonumber\\ & \quad +|\{|u_m|>\tau\}| + |\{| u_l - u_m |>r\}|+E, \end{align}

where

(5.11)\begin{align} E=|\{| u_l - u_m |\leq r,\,|u_l|\leq \tau,\,|u_m|\leq \tau,\,|\nabla u_l|\leq\tau,\,|\nabla u_m|\leq \tau,\,|\nabla u_l-\nabla u_m|>t\}|. \end{align}

Note that (4.4) and arguments of the proof of [Reference Chlebicka and Zatorska-Goldstein32, lemma 4.12] enable to justify uniform integrability of $\{g(|\nabla u_k|)\}_k$ for some continuous and increasing function $g$. Therefore, we can choose for any ${{\varepsilon }}>0$ a number $\tau _{{\varepsilon }}$ large enough so that for $\tau >\tau _{{\varepsilon }}$ we obtain

(5.12)\begin{align} |\{|\nabla u_l|>\tau\}|<{{\varepsilon}},\ |\{|\nabla u_m|>\tau\}|<{{\varepsilon}},\ |\{|u_l|>\tau\}|<{{\varepsilon}}, \ \text{and}\ |\{|u_m|>\tau\}|<{{\varepsilon}}. \end{align}

From now on we restrict ourselves to $\tau >\tau _{{\varepsilon }}$. On the other hand, since $\{u_l\}$ is a Cauchy sequence in measure,

(5.13)\begin{equation} |\{| u_l - u_m |>r\}| < {{\epsilon}}, \text{if $l,m,r$ are sufficiently large.} \end{equation}

What remains to prove is that there exists $\delta _{\tau,{{\varepsilon }}}>0$, such that for every $\delta <\delta _{\tau,{{\varepsilon }}}$, we get

(5.14)\begin{equation} |E|<{{\varepsilon}}. \end{equation}

Let us define a set

\[ S=\{(\xi,\eta)\in{{\mathbb{R}^{n}}}\times{{\mathbb{R}^{n}}}:\ |\xi|\leq\tau,\ |\eta|\leq\tau,\ |\xi-\eta|\geq t\}\,, \]

which is compact. Consider the function $\psi :\Omega \to [0,\infty )$ given by

\[ \psi(x)=\inf_{ (\xi,\eta)\in S}\left[\left(\mathcal{A}(x,\xi)-\mathcal{A}(x,\eta)\right)\cdot(\xi-\eta)\right]. \]

Monotonicity assumption ($\mathcal {A}3$) and the continuity of the function $\xi \mapsto \mathcal {A}(\cdot,\xi )$ a.e. in $S$ ensure that $\psi \geq 0$ in $\Omega$. Furthermore, ($\mathcal {A} 4$) implies that $|\{\mathcal {A}(x,0)=0\}|=0$. Moreover,

(5.15)\begin{align} \int_S\psi(x)\,{\rm d}x& \leq \int_S \left(\mathcal{A}(x,\nabla u_l)-\mathcal{A}(x,\nabla u_m)\right)\cdot (\nabla u_l-\nabla u_m)\,{\rm d}x\\ & \leq \int_{\{|u_l-u_m|\leq r\}} \left(\mathcal{A}(x,\nabla u_l)-\mathcal{A}(x,\nabla u_m)\right) \cdot (\nabla u_l-\nabla u_m)\,{\rm d}x\nonumber\\ & = \int_{\Omega} \left(\mathcal{A}(x,\nabla u_l)-\mathcal{A}(x,\nabla u_m)\right)\cdot (\nabla T_r( u_l- u_m))\,{\rm d}x\nonumber\\ & =\int_{\Omega} T_r( u_l- u_m) \,{\rm d}\mu^{s}(x)\ \leq\ 2r\|\mu\|_{\mathcal{M}_b(\Omega)} , \nonumber \end{align}

where the last but one equality follows on making use of the test function $T_r(u_l-u_m)$ and in the corresponding equation with $l$ replaced by $m$, and subtracting the resultant equations. Estimate (5.15) and the properties of the function $\psi$ ensure that, if $s$ is chosen sufficiently small, then (5.14) holds. From inequalities (5.10), (5.12), (5.14) and (5.13), we infer that $\{\nabla u_s\}_s$ is a Cauchy sequence in measure.

To conclude that the function $u$ obtained in (5.3) and (5.2) is a desired approximable solution, we observe that it belongs to the class $\mathcal {T} ^{1,\varphi (\cdot )}_0(\Omega )$, and that $\nabla u_s \to \nabla u$ $\hbox {a.e. in} \Omega$ (up to subsequences), where $\nabla u$ is understood in the sense of (1.4). Since $\{\nabla u_s\}$ is a Cauchy sequence in measure, there exist a subsequence (still indexed by $s$) and a measurable function $W:\Omega \to {{\mathbb {R}^{n}}}$ such that $\nabla u_s\to W$ a.e. in $\Omega.$ To motivate that $\nabla u = W$ and

(5.16)\begin{equation} \chi_{\{|u|< k\}} W \in (L^{\varphi({\cdot})}(\Omega))^{n} \quad \text{for every }k>0 \end{equation}

it suffices to recall (5.3). Indeed, then for each fixed $k>0$, there exists a subsequence of $\{u_s\}$, still indexed by $s$, such that

(5.17)\begin{equation} \lim _{s \to \infty} \nabla T_k(u_s)= \lim _{s \to \infty} \chi_{\{|u_s|< k\}} \nabla u_s = \chi_{\{|u|< k\}} W \quad \text{a.e. in }\Omega, \end{equation}

and $\lim _{s \to \infty } \nabla T_k(u_s) = \nabla T_k(u)$ weakly in $(L^{\varphi (\cdot )}(\Omega ))^{n}$. Therefore, $\nabla T_k(u) = \chi _{\{|u|< k\}} W$ a.e. in $\Omega$, whence (5.16) follows. Then, due to ($\mathcal {A} 1$) also (5.9) holds, that is we have (5.4) and (5.5). Due to remark 2.5 we get uniform integrability of $\{\mathcal {A}(x,\nabla T_k u)\}_k$, so Lebesgue's monotone convergence theorem justifies (5.7), where the limit is in $(L^{1}(\Omega ))^{n}$ by lemma 2.10. By (4.3) and Vitali's convergence theorem we infer (5.6).

Proof. We prove first weak-$\ast$ convergence of measures generated by truncations of solutions and then their further properties.

Step 1. $\lambda _k\in {{\mathcal {M}_b}}(\Omega )$ and ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku)\rightharpoonup {} {{\mathfrak {A}_{\varphi (\cdot )}}}(u)$ weakly- $\ast$ in the space of measures.

For $k>\delta >0$ we define a Lipschitz functions $h_\delta,\sigma _\delta ^{+},\sigma _\delta ^{-}:\mathbb {{R}}\to \mathbb {{R}}$ satisfying

\[ \begin{cases} h_\delta(r)=1 & \ \text{ if }|r|\leq k-\delta,\\ |h_\delta'(r)|=\tfrac 1\delta & \ \text{ if }k-\delta\leq |r|\leq k,\\ h_\delta(r)=0 & \ \text{ if }|r|\geq k. \end{cases}\quad \begin{cases} \sigma_\delta^{+}(r)=0 & \ \text{ if }r\leq k-\delta,\\ (\sigma_\delta^{+})'(r)=\tfrac 1\delta & \ \text{ if }k-\delta\leq r\leq k,\\ \sigma_\delta^{+}(r)=1 & \ \text{ if }r\geq k, \end{cases} \]

and $\sigma _\delta ^{-}(r)=\sigma _\delta ^{+}(-r)$. We note that if $\{u_s\}$ is an approximate sequence from definition 1.6 solving (4.1) with $\mu ^{s}$ being bounded and smooth function, $\phi \in C_0^{\infty }(\Omega )$, then $h_\delta (u_s)\phi,\sigma _\delta ^{+}(u_s),\sigma _\delta ^{-}(u_s)\in W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$ are admissible test functions in (1.6). By testing (4.1) by $h_\delta (u_s)\phi$ we get

\begin{align*} \int_\Omega h_\delta(u_s)\,\mathcal{A}(x,\nabla u_s)\cdot \nabla\phi\,{\rm d}x & =\int_\Omega \phi\mu^{s} h_\delta(u_s)\,{\rm d}x-\int_\Omega h'_\delta(u_s)\,\mathcal{A}(x,\nabla u_s)\cdot \nabla u_s\,\phi\,{\rm d}x\\ & =\int_\Omega \phi\,{\rm d}\lambda_\delta^{s}+\int_\Omega \phi\,{\rm d}\gamma_\delta^{s,+}-\int_\Omega \phi\,{\rm d}\gamma_\delta^{s,-}, \end{align*}

where

(6.5)\begin{align} \lambda_\delta^{s}& =\mu^{s}h_\delta(u_s),\nonumber\\ \nu_\delta^{s,+}& =\tfrac 1\delta\mathbb {1}_{\{k-\delta\leq u_s\leq k\}}\mathcal{A}(x,\nabla u_s)\cdot\nabla u_s, \end{align}

(6.6)\begin{align} \nu_\delta^{s,-}& =\tfrac 1\delta\mathbb {1}_{\{- k\leq u_s\leq{-}k+\delta\}}\mathcal{A}(x,\nabla u_s)\cdot\nabla u_s. \end{align}

Observe that

\[ \lambda_\delta^{s},\nu_\delta^{s,+},\nu_\delta^{s,-}\in L^{1}(\Omega). \]

Indeed,

\[ \|\lambda_\delta^{s}\|_{L^{1}(\Omega)}\leq\int_\Omega|\mu^{s}|\,|h_\delta(u_s)|\,{\rm d}x\leq \int_\Omega|\mu^{s}| \,{\rm d}x\leq 2\|\mu\|_{{{\mathcal{M}_b}}(\Omega)}. \]

To estimate $\|\nu _\delta ^{s,+}\|_{L^{1}(\Omega )}$ and $\|\nu _\delta ^{s,-}\|_{L^{1}(\Omega )}$, we test (4.1) against $\sigma _\delta ^{+}(u_s)$ (respectively $\sigma _\delta ^{-}(u_s)$) and obtain

(6.7)\begin{align} \|\nu_\delta^{s,+}\|_{L^{1}(\Omega)}& \leq \frac 1\delta \int_{\{k-\delta\leq u_s\leq k\}}\mathcal{A}(x,\nabla u_s)\cdot\nabla u_s\,{\rm d}x\notag\\ & =\int_\Omega \mu^{s}\sigma^{+}_\delta(u_s)\,{\rm d}x\leq 2\|\mu\|_{{{\mathcal{M}_b}}(\Omega)}, \end{align}

(6.8)\begin{align} \|\nu_\delta^{s,-}\|_{L^{1}(\Omega)}& \leq \frac 1\delta \int_{\{{-}k\leq u_s\leq{-}k+\delta\}}\mathcal{A}(x,\nabla u_s)\cdot\nabla u_s\,{\rm d}x\notag\\ & =\int_\Omega \mu^{s}\sigma^{-}_\delta(u_s)\,{\rm d}x\leq 2\|\mu\|_{{{\mathcal{M}_b}}(\Omega)}. \end{align}

In the end we have that

\[ \left\|-{{\rm div\,}} \Big(h_\delta(u_s)\,\mathcal{A}(x,\nabla u_s)\Big)\right\|_{L^{1}(\Omega)}\leq 6\|\mu\|_{{{\mathcal{M}_b}}(\Omega)}. \]

Due to remark 2.5 and ($\mathcal {A}$2) we get uniform integrability of $\{\mathcal {A}(x,\nabla (T_k (u_s)))\}_k$, so Lebesgue's monotone convergence theorem justifies we can let $\delta \to 0$ getting

\[ |h_\delta(u_s)\,\mathcal{A}(x,\nabla u_s)| \to |\mathcal{A}(x,\nabla( T_k( u_s)))|\quad \text{strongly in }\ L^{1}(\Omega). \]

Therefore ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku_s)\in {{\mathcal {M}_b}}(\Omega )$ and $\|{{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku_s) \|_{{{\mathcal {M}_b}}(\Omega )}\leq 6\|\mu \|_{{{\mathcal {M}_b}}(\Omega )}$, where the bound is uniform with respect to $s$ and $k$. Consequently, the use of proposition 5.1 enables to infer that also that ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku)\in {{\mathcal {M}_b}}(\Omega )$, $\|{{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku) \|_{{{\mathcal {M}_b}}(\Omega )}\leq 6\|\mu \|_{{{\mathcal {M}_b}}(\Omega )}$, and – finally – (6.2). By remark 2.4 we can extend the family of admissible test functions to get (6.1) and conclusion that ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku)\in {{\mathcal {M}_b}}(\Omega )\cap (W^{1,\varphi (\cdot )}_0(\Omega ))'$.

Step 2. Existence of a diffuse measure $\vartheta \in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$, such that

\[ \vartheta {\mathsf {L}} \{|u|< k\}= {{\mathfrak{A}_{\varphi({\cdot})}}}(T_{l}u) {\mathsf {L}} \{|u|< k\}\text{ for every }k>0\text{ and every }{l}\geq k. \]

Lemma 2.10 ensures that $\phi \in W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$ belongs to $L^{1}(\Omega,\vartheta )$ with any $\vartheta \in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$.

Note that $\lambda _{l} {\mathsf {L}} \{|u|< k\}=\lambda _k {\mathsf {L}} \{|u|< k\}$ for every ${l} \geq k>0$. Since the set $\{|u|< k\}$ is ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-open, lemmas 2.8 and 2.9 ensure that there exists an increasing sequence $\{w_j\}$ of nonnegative functions in $W^{1,\varphi (\cdot )}_0(\Omega )$ which converges to $\mathbb {1}_{\{|u|< k\}}$ ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-everywhere in $\Omega$. Then $w_j=0$ a.e. in $\{|u|\geq k\}$. If $\psi \in C_0^{\infty }$, then $\phi =w_j\psi \in W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$ is an admissible test function in (6.1), so for every ${l} \geq k$ we get

\begin{align*} \int_\Omega w_j\psi\,{\rm d}\lambda_k& =\int_{\{|u|\leq k\}} \mathcal{A}(x,\nabla u)\cdot\nabla( w_j\psi)\,{\rm d}x\\ & =\int_{\{|u|\leq {l}\}} \mathcal{A}(x,\nabla u)\cdot\nabla( w_j\psi)\,{\rm d}x=\int_\Omega w_j\psi\,{\rm d}\lambda_{l}. \end{align*}

Passing to the limit with $j\to \infty$ we get

\[ \int_{\{|u|< k\}} \psi\,{\rm d}\lambda_k= \int_{\{|u|< k\}} \psi\,{\rm d}\lambda_{l}\quad\text{ for every }\psi\in C_0^{\infty}, \]

so of course $\lambda _{l} {\mathsf {L}} \{|u|< k\}=\lambda _k {\mathsf {L}} \{|u|< k\}$. Consequently, there exists a unique (up to sets of $W^{1,\varphi (\cdot )}$-capacity zero) Borel measure $\vartheta$, such that $\vartheta {\mathsf {L}} \{|u|=+\infty \}=0$ and $\vartheta {\mathsf {L}} \{|u|< k\}=\lambda _{l} {\mathsf {L}} \{|u|< k\}$ for every $k>0$ and every ${l} \geq k$. As $\lambda _k$ vanishes on every set of zero capacity ${{{\rm C}_{\varphi (\cdot )}}}$, so does $\vartheta$. By (6.2) the measures $|\lambda _k|$ are uniformly bounded with respect to $k$, so $\{|\vartheta |(\{|u|< k\})\}_k$ is bounded. In turn, $|\vartheta |(\Omega )<\infty$ and – finally – we infer that $\vartheta \in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$.

Step 3. ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku) {\mathsf {L}} \{|u|>k\}=0$.

Lemma 2.7 gives that $u$ is ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-continuous, thus the set $\{|u|>k\}$ is ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-open. Fix arbitrary open $V\subset \Omega$. By lemma 2.9, there exists an increasing sequence $\{\widehat w_j\}$ of nonnegative functions in $W^{1,\varphi (\cdot )}_0(\Omega )$ which converges to $\mathbb {1}_{V\cap \{|u|< k\}}$ ${{{\rm C}_{\varphi (\cdot )}}}$-quasi-everywhere in $\Omega$. Then $\widehat w_j=0$ a.e. in $\{|u|\leq k\}$ and we can test (6.1) against $w_j \in W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$. We obtain

\[ \int_\Omega w_j\,{\rm d}\lambda_k=\int_{\{|u|\leq k\}} \mathcal{A}(x,\nabla u)\cdot\nabla( w_j )\,{\rm d}x=0. \]

Letting $j\to \infty$ we get that $(\lambda _k {\mathsf {L}} \{|u|>k\})(V)=0$. Since $V$ was arbitrary open set, we have what was claimed.

Step 4. Limits. Since we have (6.7) and (6.8), we get (6.3) and (6.4) for any $\phi \in C_0(\Omega )$, with some nonnegative $\nu _k^{+},\nu _k^{-}\in {{\mathcal {M}_b}}(\Omega )$. They have the form given in the claim, because ${{\mathfrak {A}_{\varphi (\cdot )}}}(T_ku)\in {{\mathcal {M}_b}}(\Omega )\cap (W^{1,\varphi (\cdot )}_0(\Omega ))'$ has properties proven in steps 3 and 4.

Proof. By proposition 5.1 there exists an approximable solution $u\in \mathcal {T}^{1,\varphi (\cdot )}_0(\Omega )$ to (1.1). We shall show that actually it is also a renormalized solution. Due to proposition 6.1, measure $\mu$ can be seen as the weak-$\ast$ limit of $\{\lambda _k\}$, which are expressed as

\[ \lambda_k={{\mathfrak{A}_{\varphi({\cdot})}}}(T_ku)=\vartheta {\mathsf {L}} \{|u|< k\}+\nu_k^{+}-\nu_k^{-} \]

with $\vartheta \in {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega )$, $\nu _k^{+},\nu _k^{-}\in ( {{\mathcal {M}_b}}(\Omega )\setminus {{\mathcal {M}^{\varphi (\cdot )}_b}}(\Omega ))\cup \{0\}$ being such that $\nu _k^{+}=\nu _k^{+} {\mathsf {L}} \{u=k\}$ and $\nu _k^{-}=\nu _k^{-} {\mathsf {L}} \{u=-k\}$. Given $h\in W^{1,\infty }(\mathbb {{R}})$ having $h'$ with compact support, $\phi \in C_0^{\infty }(\Omega )$, and arbitrary $k>0$, function $h(T_{k+1}(u))\phi \in W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$, so we can test equation (6.1) to get

(6.9)\begin{align} & \int_{\{|u|\leq k\}}\mathcal{A}(x,\nabla u)\cdot\nabla \big(h(T_{k+1}(u))\phi\big)\,{\rm d}x= \int_{\{|u|\leq k+1\}}h(u)\phi\,{\rm d}\lambda_k \end{align}

(6.10)\begin{align} & \quad=\int_{\{|u|< k\}}h(u)\phi\,{\rm d}\vartheta+h(k)\int_\Omega \phi\,{\rm d}\nu_k^{+}-h({-}k)\int_\Omega \phi\,{\rm d}\nu_k^{-}. \end{align}

We need to justify letting $k\to \infty$. We start with the left-hand side of (6.9) by having a look on

\[ \mathcal{A}(x,\nabla u)\cdot \nabla \big(h(u)\phi\big)=\mathcal{A}(x,\nabla u)\cdot\nabla u\,(h'(u)\phi)+\mathcal{A}(x,\nabla u)\cdot\nabla \phi\, h(u). \]

If we prove that both terms on the right-hand side in the last display are integrable, Lebesgue's monotone convergence theorem will give the desired conclusion. Recall that $u\in \mathcal {T}^{1,\varphi (\cdot )}_0(\Omega )$ and satisfy (4.2), so by proposition 4.2 and lemma 2.10, $\mathcal {A}(\cdot,\nabla u)\in (L^{1}(\Omega ))^{n}$. Moreover, $h'$ is bounded and ${\rm supp}\,h'\subset [-M,M]$ for some $M>0$, so

\[ \mathcal{A}({\cdot},\nabla u)\cdot\nabla u\,h'(u)=\mathcal{A}({\cdot},\nabla T_Mu)\cdot\nabla(T_M u)\,h'(u) \]

is integrable by (4.2). For the second term we see that

\[ \|\mathcal{A}(x,\nabla u)\cdot\nabla \phi\, h(u)\|_{L^{1}(\Omega)}\leq \|\mathcal{A}(x,\nabla u)\|_{L^{1}(\Omega)}\,\|\nabla \phi\|_{L^{\infty}(\Omega)}\,\|h\|_{L^{\infty}(\Omega)}, \]

so it suffices to use the same arguments as before. Therefore, (6.9) becomes the left-hand side of (1.8) in the limit. By remark 1.3 the following decomposition

\begin{align*} & \mu={{\mu_{\varphi({\cdot})}}}+{{\mu^{+}_{\rm sing}}}-{{\mu^{-}_{\rm sing}}},\quad {{\mu_{\varphi({\cdot})}}}\in{{\mathcal{M}^{\varphi({\cdot})}_b}}(\Omega),\ 0\leq {{\mu^{+}_{\rm sing}}},\\ & {{\mu^{-}_{\rm sing}}}\in\big( {{\mathcal{M}_b}}(\Omega)\setminus{{\mathcal{M}^{\varphi({\cdot})}_b}}(\Omega)\big)\cup\{0\} \end{align*}

is unique up to sets of $W^{1,\varphi (\cdot )}$-capacity zero. By (6.2) it holds that $\vartheta {\mathsf {L}} \{|u|< k\}\rightharpoonup {{\mathfrak {A}_{\varphi (\cdot )}}}(u)$. Note that it is also (6.2) to justify testing against $W^{1,\varphi (\cdot )}_0(\Omega )\cap L^{\infty }(\Omega )$-function. To conclude we use Lebesgue's dominated convergence theorem in (6.10). To motivate the convergence of the first term we note that we can split the first term to positive and negative part, whose majorants are integrable due to lemma 2.10. For the remaining two terms it suffices to recall that $h$ is bounded and constant in infinities. By (6.3) one has $\nu _k^{+}\rightharpoonup {{\mu ^{+}_{\rm sing}}}$ with ${\rm supp}\,{{\mu ^{+}_{\rm sing}}}\subset \cap _{k>0}\{u>k\}$, and by (6.4) also $\nu _k^{-}\rightharpoonup {{\mu ^{-}_{\rm sing}}}$ with ${\rm supp}\,{{\mu ^{-}_{\rm sing}}}\subset \cap _{k>0}\{u<-k\}$.

Proof. We note first that the problem is well posed as $\mu ^{i,s}=f^{i}_s-{{\rm div\,}}G^{i}_s\rightharpoonup \mu _{\varphi (\cdot )}$ weakly-$*$ in the space of measures, $i=1,2$. We fix arbitrary $t,l>0$, use $\phi =T_t(T_l(v^{1}_s)-T_l(v^{2}_s))\in W_0^{1,\varphi (\cdot )}(\Omega )\cap L^{\infty }(\Omega )$ as a test function in both (1.6) and subtract the equations to obtain for every $s>0$

(7.4)\begin{align} L_s=& \int_{\{|T_l(v^{1}_s)-T_l(v^{2}_s)|\leq t\}}(\mathcal{A}(x,\nabla v^{1}_s)- \mathcal{A}(x,\nabla v^{2}_s) )\cdot( \nabla v^{1}_s-\nabla v^{2}_s)\,{\rm d}x\nonumber\\ =& \int_\Omega(f^{1}_s-f^{2}_s)T_t(T_l(v^{1}_s)-T_l(v^{2}_s))\,{\rm d}x\notag\\ & \quad +\int_\Omega (G^{1}_s-G^{2}_s)\cdot\nabla T_t(T_l(v^{1}_s)-T_l(v^{2}_s))\,{\rm d}x = R_s^{1}+R_s^{2}. \end{align}

The right-hand side above tends to $0$. Indeed, the convergence of $R_s^{1}$ holds because $|T_t(T_lv^{1}_s-T_lv^{2}_s)|\leq t$ and for $s\to 0$ we have $f^{1}_s-f^{2}_s\to 0$ in $L^{1}(\Omega )$. As for $R_s^{2}$ it suffices to note that

\begin{align*} |R_s^{2}|& =\left|\int_{\{|T_l(v^{1}_s)-T_l(v^{2}_s)|\leq t\}}(G^{1}_s-G^{2}_s)\cdot\nabla T_lv^{1}_s\,{\rm d}x\right.\\ & \quad -\left.\int_{\{|T_l(v^{1}_s)-T_l(v^{2}_s)|\leq t\}}(G^{1}_s-G^{2}_s)\cdot\nabla T_l(v^{2}_s)\,{\rm d}x\right|\\ & \leq \left|\int_\Omega(G^{1}_s-G^{2}_s)\cdot\nabla T_l(v^{1}_s)\,{\rm d}x\right|+\left|\int_\Omega (G^{1}_s-G^{2}_s)\cdot\nabla T_l(v^{2}_s) \,{\rm d}x\right|\\ & \leq 2\| G^{1}_s-G^{2}_s\|_{(L^{\widetilde\varphi({\cdot})}(\Omega))^{n}}\|\nabla T_l(v^{1}_s)\|_{(L^{\varphi({\cdot})}(\Omega))^{n}}\\ & \quad +2\| G^{1}_s-G^{2}_s\|_{(L^{\widetilde\varphi({\cdot})}(\Omega))^{n}}\|\nabla T_l(v^{2}_s)\|_{(L^{\varphi({\cdot})}(\Omega))^{n}}\\ & \leq c\| G^{1}_s-G^{2}_s\|_{(L^{\widetilde\varphi({\cdot})}(\Omega))^{n}}, \end{align*}

where we used that weak convergence of the $\{\nabla T_l(v^{j}_s)\}_s$ $(j=1,2)$ in $(L^{\varphi (\cdot )}(\Omega ))^{n}$, which in particular implies uniform boundedness of $\{\|\nabla T_l(v^{j}_s)\|_{L^{\varphi (\cdot )}(\Omega )}\}_s$ $(j=1,2)$ and recalled that the strong convergence of $(G^{1}_s-G^{2}_s)\to 0$ in $(L^{\widetilde \varphi (\cdot )}(\Omega ))^{n}$. The left-hand side of (7.4) is nonnegative due to the monotonicity of $\mathcal {A}$. Moreover, as $R_s^{1}+R_s^{2}\to 0$, we get

\begin{align*} 0\leq & \int_{\{|T_lv^{1}-T_lv^{2} |\leq t\}}(\mathcal{A}(x,\nabla v^{1})-\mathcal{A}(x,\nabla v^{2} ))\cdot( \nabla v^{1} -\nabla v^{2} )\,{\rm d}x\\ \leq & \limsup_{s\to 0}L_s= \limsup_{s\to 0}\,(R_s^{1}+R_s^{2})= 0. \end{align*}

Consequently, $\nabla v^{1} =\nabla v^{2}$ a.e. in $\{|T_l( v^{1}) -T_l( v^{2}) |\leq t\}$ for every $t,l>0$, and so

(7.5)\begin{equation} \nabla v^{1} =\nabla v^{2}\quad\text{ a.e. in }\Omega. \end{equation}

Given the boundary value also $v^{1}= v^{2}$ a.e. in $\Omega$.

Proof of theorem 1.8. Existence of approximable solutions is provided in proposition 5.1. Proposition 6.2 yields that an approximable solution is a renormalized solutions. Proposition 6.1 actually localizes the support of singular measures. Approximable solutions can be achieved from renormalized ones by a choice of $h=T_k$.