$\mathcal{H}^{1}(J_{\widetilde{u}_{h}})+\int_{\Omega}|e(\widetilde{u}_{h})|^{2}dx\leq\mathcal{F}(A_{m_{h}},u_{m_{h}})+\mathcal{H}^{1}(\Sigma)\leq
M:=\mathcal{F}(A_{1},u_{1})+\mathcal{H}^{1}(\Sigma)<\infty$ (4.20)

for every $h\geq 1.$ Moreover, by Proposition 4.4, the definitions of $u,$
$\widetilde{u}_{h}$ and $\widetilde{u},$

$\widetilde{u}_{h}\to\widetilde{u}\quad\text{a.e. in $[S\cup(\Omega\setminus
G)]\cup\bigcup\limits_{i\in N}E_{i}$}$ (4.21)

and

$|\widetilde{u}_{h}|\to+\infty\quad\text{a.e. in
$\bigcup\limits_{j\in\mathbb{N}\setminus N}E_{j},$}$ (4.22)

where $\\{E_{i}\\}$ and $N$ are provided by Proposition 4.4.

We observe that a priori $\partial A,$ and hence $\partial G$, does not need
to be $\mathcal{H}^{1}$-rectifiable. Therefore, by [22, Theorem 6.2]
$J_{\widetilde{u}}\subset(\Omega\cap\partial^{r}G)\cup(\Sigma\cap J_{u}),$
where we recall that $\partial^{r}G$ is $\mathcal{H}^{1}$-rectifiable part of
$\partial G.$

To prove (4.13) we use similar arguments as in [46, Proposition 4.1]. Let
$g\in L^{\infty}(\Sigma\times\\{0,1\\})$ be such that

$g(x,s):=\varphi(x,\nu_{\Sigma}(x))+s\beta(x)$

for which we know by (2.13) that $g\geq 0$ and

$|g(x,1)-g(x,0)|\leq\varphi(x,\nu_{\Sigma}(x))\qquad\text{for a.e.
$x\in\Sigma.$ }$ (4.23)

Let $\mu_{h}$ be the sequence of positive Radon measures defined at Borel sets
$B\subset\mathbb{R}^{2}$ as

$\displaystyle\mu_{h}(B)$
$\displaystyle:=\int_{B\cap\Omega\cap\partial^{*}A_{m_{h}}}\varphi(x,\nu_{A_{m_{h}}})d\mathcal{H}^{1}+2\int_{B\cap\Omega\cap(A_{m_{h}}^{(1)}\cup
A_{m_{h}}^{(0)})\cap\partial
A_{m_{h}}}\varphi(x,\nu_{A_{m_{h}}})d\mathcal{H}^{1}$
$\displaystyle+\int_{B\cap\Sigma\cap A_{m_{h}}^{(0)}\cap\partial
A_{m_{h}}}\big{[}\varphi(x,\nu_{\Sigma})+g(x,1)\big{]}d\mathcal{H}^{1}(x)+\int_{B\cap\Sigma\setminus\partial
A_{m_{h}}}g(x,0)d\mathcal{H}^{1}$
$\displaystyle+\int_{B\cap\Sigma\cap\partial^{*}A_{m_{h}}\setminus
J_{u_{m_{h}}}}g(x,1)d\mathcal{H}^{1}+\int_{B\cap\Sigma\cap
J_{u_{m_{h}}}}\big{[}g(x,0)+\varphi(x,\nu_{\Sigma})\big{]}\,d\mathcal{H}^{1}$

and let $\mu$ be the positive measure defined at Borel sets
$B\subset\mathbb{R}^{2}$ as

$\displaystyle\mu(B):=\int_{B\cap\Omega\cap\partial^{*}G}\varphi(x,\nu_{A})d\mathcal{H}^{1}+2\int_{B\cap\Omega\cap
G^{(1)}\cap\partial G\cap
J_{\widetilde{u}}}\varphi(x,\nu_{G})d\mathcal{H}^{1}$ $\displaystyle+$
$\displaystyle\int_{B\cap\Sigma\setminus\partial
G}g(x,0)d\mathcal{H}^{1}+\int_{B\cap\Sigma\cap\partial^{*}G\setminus
J_{u}}g(x,1)d\mathcal{H}^{1}+\int_{B\cap\Sigma\cap
J_{\widetilde{u}}}\big{[}g(x,0)+\varphi(x,\nu_{\Sigma})\big{]}\,d\mathcal{H}^{1}.$

Since $S_{\widetilde{u}}^{A}:=G^{(1)}\cap\partial G\cap J_{\widetilde{u}}$ and
$\Sigma\cap J_{\widetilde{u}}=\Sigma\cap J_{u},$ we have

$\mu_{h}(\mathbb{R}^{2})=\mathcal{S}(A_{m_{h}},u_{m_{h}})+\int_{\Sigma}\varphi(x,\nu_{\Sigma})d\mathcal{H}^{1}$

and

$\mu(\mathbb{R}^{2})=\widetilde{\mathcal{S}}(G,u)+\int_{\Sigma}\varphi(x,\nu_{\Sigma})d\mathcal{H}^{1}.$

Hence, to establish (4.13) it suffices to prove

$\liminf\limits_{h\to\infty}\mu_{h}(\mathbb{R}^{2})\geq\mu(\mathbb{R}^{2}).$
(4.24)

Since $\sup_{h}\mu_{h}(\mathbb{R}^{2})<+\infty,$ by compactness, there exists
a positive Radon measure $\mu_{0}$ in $\mathbb{R}^{2}$ such that (up to a
subsequence) $\mu_{h}\rightharpoonup^{*}\mu_{0}$ as $h\to\infty.$ We show

$\mu_{0}\geq\mu$ (4.25)

and we observe that (4.24) immediately follows from (4.25). To establish
(4.25) it suffices to prove

$\displaystyle\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(\Omega\cap\partial^{*}G)}(x)\geq\varphi(x,\nu_{G}(x))\qquad\text{for
a.e. $x\in\Omega\cap\partial^{*}G,$}$ (4.26a)
$\displaystyle\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(\Sigma\cap\partial^{*}G)}(x)\geq
g(x,1)\qquad\text{for a.e. $x\in\Sigma\cap\partial^{*}G,$}$ (4.26b)
$\displaystyle\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(\Sigma\setminus\partial
G)}(x)\geq\varphi(x,\nu_{\Sigma}(x))\qquad\text{for a.e.
$x\in\Sigma\setminus\partial G,$}$ (4.26c)
$\displaystyle\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits S_{\widetilde{u}}^{A}}(x)\geq
2\varphi(x,\nu_{G}(x))\qquad\text{for a.e. $x\in S_{\widetilde{u}}^{A},$ }$
(4.26d) $\displaystyle\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(\Sigma\cap
J_{\widetilde{u}})}(x)\geq 2\varphi(x,\nu_{\Sigma}(x))\qquad\text{for a.e.
$x\in\Sigma\cap J_{\widetilde{u}}$}$ (4.26e)

since $g(x,0)=\varphi(x,\nu_{\Sigma}).$ We use similar arguments of the proof
of [46, Proposition 4.1].

The proof of the estimates (4.26a)-(4.26e) follows from similar arguments used
in [46, Proposition 4.1] with special care needed for (4.26d). In fact for
(4.26d) we cannot employ the strategy used for [46, Eq. 4.40c] which was
hinged on the uniform bound on the number of boundary components, which here
we do not have. We instead adapt the arguments employed in [46, Eq. 4.40g] by
using Lemma 4.7.

Next we detail the proofs of (4.26a)-(4.26e).

Proof of (4.26a). Note that $A=G$ up to a negligible set. By Corollary 4.2
$A_{m_{h}}\to A$ in $L^{1}(\mathbb{R}^{2}),$ thus, the proof of (4.26a) can be
done following the arguments of [46, Eq. 4.40a] using Reshetnyak lower
semicontinuity Theorem [3, Theorem 2.38].

Proof of (4.26b). Since $A_{m_{h}}\to G$ in $L^{1}(\mathbb{R}^{2}),$ we have
$D\chi_{A_{m_{h}}}\rightharpoonup^{*}D\chi_{G}.$ Thus, the proof of (4.26b)
directly follows from [2, Lemma 3.8] (see also the proof of [46, Eq. 4.40d]).

Proof of (4.26c). Let $x_{0}\in\Sigma\setminus\partial G$ and let
$r_{0}:=\mathrm{dist}(x_{0},\partial G)>0.$ Since
$\mathbb{R}^{2}\setminus\mathrm{Int}(A_{m_{h}})\overset{K}{\to}\mathbb{R}^{2}\setminus\mathrm{Int}(A)=\mathbb{R}^{2}\setminus
G,$ there exists $r_{1}\in(0,r_{0})$ such that
$B_{r}(x_{0})\cap\mathrm{Int}(A_{m_{h}})$ for any $r\in(0,r_{1}).$ Therefore,
for any $r\in(0,r_{1})$

$\displaystyle\mu_{h}(\overline{B_{r}(x_{0})})=$
$\displaystyle\int_{\overline{B_{r}(x_{0})}\cap\Sigma\cap
A_{m_{h}}^{(0)}\cap\partial
A_{m_{h}}}\big{[}\varphi(x,\nu_{\Sigma})+g(x,1)\big{]}d\mathcal{H}^{1}(x)+\int_{\overline{B_{r}(x_{0})}\cap\Sigma\setminus\partial
A_{m_{h}}}g(x,0)d\mathcal{H}^{1}$ $\displaystyle\geq$
$\displaystyle\int_{\overline{B_{r}(x_{0})}\cap\Sigma\cap
A_{m_{h}}^{(0)}\cap\partial
A_{m_{h}}}g(x,0)d\mathcal{H}^{1}(x)+\int_{\overline{B_{r}(x_{0})}\cap\Sigma\setminus\partial
A_{m_{h}}}g(x,0)d\mathcal{H}^{1}$ $\displaystyle=$
$\displaystyle\int_{\overline{B_{r}(x_{0})}\cap\Sigma}g(x,0)d\mathcal{H}^{1},$

where in the inequality we used (4.23). Thus, taking $limsup$ as $h\to+\infty$
and using $\mu_{h}\rightharpoonup^{*}\mu_{0}$ we get

$\mu_{0}(\overline{B_{r}(x_{0})})\geq\int_{B_{r}(x_{0})\cap\Sigma}g(x,0)d\mathcal{H}^{1}.$

Now (4.26c) follows from the Besicovitch Differentiation Theorem.

Proofs of (4.26d) and (4.26e). We establish

$\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits K}\geq
2\phi(x,\nu_{K})\quad\text{for $\mathcal{H}^{1}$-a.e. $x\in K,$}$ (4.27)

where

$K=S_{\widetilde{u}}^{A}\cup(\Sigma\cap J_{\widetilde{u}}).$

Let $x\in K$ be such that $\theta(K,x)=1.$ Then either $x\in
S_{\widetilde{u}}\subset G^{(1)}\cap\partial^{r}G$ or $x\in\Sigma\cap
J_{\widetilde{u}}.$ By setting $E_{0}:=S$ and recallling that
$\mathrm{Int}(A)=\cup_{i\in\mathbb{N}}E_{i},$ in view of Proposition 4.4 we
have the one of the following:

  * (b1)

there exists $i_{0}\in N$ such that $x\in
E_{i_{0}}^{(1)}\cap\partial^{r}E_{i_{0}},$
$\theta(E_{i_{0}}^{(1)}\cap\partial^{r}E_{i_{0}},x)=1$ and
$u_{m_{h}}+a_{m_{h}}\to u$ a.e. in $E_{i_{0}};$

  * (b2)

there exist $i_{0}\in N\cup\\{0\\}$ and $j_{0}\in\mathbb{N}\setminus N$ such
that $x\in\partial^{*}E_{i_{0}}\cap\partial^{*}E_{j_{0}}$ and
$u_{m_{h}}+a_{m_{h}}\to u$ a.e. in $E_{i_{0}}$ and
$|u_{m_{h}}+a_{m_{h}}|\to\infty$ a.e. in $E_{j_{0}};$

  * (b3)

there exist $i_{1},i_{2}\in N\cup\\{0\\}$ with $i_{1}\neq i_{2}$ such that
$x\in\partial^{*}E_{i_{1}}\cap\partial^{*}E_{i_{2}}$ and
$u_{m_{h}}+a_{m_{h}}\to u$ a.e. in $E_{i_{1}}\cup E_{i_{2}}.$

Let $L$ denote the set among $E_{i_{0}}^{(1)}\cap\partial^{r}E_{i_{0}},$
$\partial^{*}E_{i_{0}}\cap\partial^{*}E_{j_{0}}$ and
$\partial^{*}E_{i_{1}}\cap\partial^{*}E_{i_{2}}$ containing $x.$ Without loss
of generality we assume that $x\in Y\subset L,$ where $Y$ is defined as the
set of points $y\in L\subset\partial A$ satisfying:

  * (c1)

$\theta(\partial G,y)=\theta(\partial A,y)=\theta(L,y)=1$ and
$\nu_{G}(y)=\nu_{A}(y)=\nu_{L}(y)$ exists. If $y\in\Sigma,$ then additionally,
$\theta(\Sigma,x)=1$ and $\nu_{\Sigma}$ also exists;

  * (c2)

as $\rho\to 0$ the sets
$\overline{Q_{R,\nu_{L}}(y)}\cap\sigma_{\rho,x}(\partial A),$
$\overline{Q_{R,\nu_{L}}(x)}\cap\sigma_{\rho,y}(\partial G)$ and
$\overline{Q_{R,\nu_{L}}(y)}\cap\sigma_{\rho,y}(L)$ converge
$\overline{Q_{R,\nu_{L}}(y)}\cap T_{y}$ in the Kuratowski sense, where $R>0$
and $T_{y}$ is the generalized tangent line to $\partial A$ at $y;$

  * (c3)

one-sided traces $\widetilde{u}^{+}(y)$ and $\widetilde{u}^{-}(y)$ of
$\widetilde{u}$ w.r.t. $L$ exist and are not equal;

  * (c4)

$\frac{d\mu_{0}}{d\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits K}\,(y)$ exists and is finite.

In fact, $\mathcal{H}^{1}(L\setminus Y)=0$ since for (c1) we notice that
$Y\subset L\subset\partial^{r}A$ and $\partial^{r}A$ is
$\mathcal{H}^{1}$-rectifiable, for (c2) we use Proposition 4.3 by observing
that the points of $\Sigma$ and $\Omega\cap\partial A$ satisfy uniform density
estimates in view of the Lipchitzianity of $\Sigma$ and Proposition 4.1,
respectively, for (c3) we use [22, Definition 2.4] and the existence of traces
of $GBD$-functions along $C^{1}$-manifolds [22, Theorem 6.2] and the fact that
being a jump set of $\widetilde{u},$ the set $K$ (and also $L$) can be covered
by at most countably many one-dimensional $C^{1}$-graphs (up to a
$\mathcal{H}^{1}$-negligible set), and finally for (c4) we use Besicovitch
Differentiation Theorem.

Without loss of generality, we assume $x=0,$ $\nu_{K}(x)={\bf e_{2}},$ $T_{x}$
is the $x_{1}$-axis and ${\bf e_{2}}$ is the outer normal of $E_{i_{0}}$.

Let $4r_{0}:=\mathrm{dist}(0,\partial\Omega)$ if $0\in\Omega$ and
$4r_{0}:=\mathrm{dist}(0,\partial\Sigma)$ if $0\in\Sigma;$ since $\Sigma$ is
Lipschitz, it consists of at most countably many open connected components in
$\partial\Omega,$ and hence, $r_{0}>0$. By weak convergence,

$\lim\limits_{h\to\infty}\mu_{h}(\overline{Q_{r}})=\mu_{0}(Q_{r})$ (4.28)

for a.e. $r\in(0,r_{0}).$ By assumption (b3), [22, Definition 2.4] and [22,
Remark 2.2] separately applied to $Q_{1}^{+}:=Q_{1}\cap\\{x_{2}>0\\}$ and
$Q_{1}\setminus Q_{1}^{+}$ we have

$\lim\limits_{r\to
0}\int_{Q_{1}}|\tau(\widetilde{u}(rx))-\tau(u_{0}(x))|dx=0,$ (4.29)

where

$u_{0}:=\widetilde{u}^{+}(0)\chi_{Q_{1}^{+}}+\widetilde{u}^{-}(0)\chi_{Q_{1}\setminus
Q_{1}^{+}}$

and

$\tau(z)=(\arctan z_{1},\arctan z_{2}),\quad
z=(z_{1},z_{2})\in\mathbb{R}^{2}.$

For every $r\in(0,r_{0})$ let

$U_{r}^{\infty}:=\\{x\in
Q_{1}:\,\,\liminf\limits_{h\to\infty}|\widetilde{u}_{h}(rx)|=+\infty\\}.$

Unlike the proof of [46, Eq. 4.40g], (4.22) implies that $U_{r}^{\infty}$ can
have positive measure. By (4.21) and the Dominated Convergence Theorem

$\lim\limits_{h\to\infty}\int_{Q_{1}\setminus
U_{r}^{\infty}}|\tau(\widetilde{u}_{h}(rx))-\tau(\widetilde{u}(rx))|dx=0.$
(4.30)

By (c2) applied with $R=8$, Proposition 4.3 and (c1)-(c3)

$\displaystyle Q_{8}\cap\sigma_{r}(\partial
A)\overset{K}{\to}I_{8}\qquad\text{and}\qquad\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(Q_{8}\cap\sigma_{r}(\partial
A))\overset{*}{\rightharpoonup}\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits I_{8},$ $\displaystyle
Q_{8}\cap\sigma_{r}(L)\overset{K}{\to}I_{8}\quad\text{and}\quad\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits(Q_{8}\cap\sigma_{r}(L)))\overset{*}{\rightharpoonup}\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits I_{8}$

as $r\to 0.$ Hence, by [46, Proposition A.5]

$\displaystyle\mathrm{sdist}(\cdot,\sigma_{r}(\partial
A))\to-\mathrm{dist}(\cdot,T_{0}),$ (4.31a)
$\displaystyle\mathrm{sdist}(\cdot,\sigma_{r}(\partial
E_{i_{0}}))\to-\mathrm{dist}(\cdot,T_{0}),$ (4.31b)
$\displaystyle\mathrm{sdist}(\cdot,\sigma_{r}(\partial[E_{i_{0}}\cup
E_{j_{0}}]))\to-\mathrm{dist}(\cdot,T_{0}),$ (4.31c)
$\displaystyle\mathrm{sdist}(\cdot,\sigma_{r}(\partial[E_{i_{1}}\cup
E_{i_{2}}]))\to-\mathrm{dist}(\cdot,T_{0})$ (4.31d)

locally uniformly in $\overline{Q_{4}}$ as $r\to 0.$ Let

$U_{0}^{\infty}=\begin{cases}\emptyset&\text{in cases (c1) and (c3)},\\\
Q_{1}^{+}&\text{in case (c2)}.\end{cases}$

By the definitions of $E_{i_{0}},$ $E_{j_{0}},$ $E_{i_{1}}$ and $E_{i_{2}}$
and (4.31b)-(4.31d)

$\lim\limits_{r\to 0}|U_{r}^{\infty}\Delta U_{0}^{\infty}|=0.$ (4.32)

Step 1: We choose sequences $h_{k}\nearrow\infty$ and $r_{k}\searrow 0$ as
follows. By (4.28), (4.29), (4.31a) and (4.32) for any $k\in\mathbb{N}$ there
exists $r_{k}\in(0,\frac{1}{k})$ such that (4.28) holds with $r=r_{k}$ and

$\displaystyle\|\mathrm{sdist}(\cdot,\sigma_{r_{k}}(\partial
A))+\mathrm{dist}(\cdot,T_{0})\|_{L^{\infty}(Q_{4})}<\frac{1}{k^{2}},$ (4.33a)
$\displaystyle\int_{Q_{1}}|\tau(\widetilde{u}(r_{k}x))-\tau(u_{0}(x))|dx<\frac{1}{k^{2}},$
(4.33b) $\displaystyle|U_{r_{k}}^{\infty}\Delta
U_{0}^{\infty}|<\frac{1}{k^{2}}.$ (4.33c)

Given $k\geq 1$ and $r_{k}$, since $A_{m_{h}}$ $sdist$-converges to $A$ and
the function $\tau$ is bounded, by (4.30), (4.33c) and (4.28) we can choose
$h_{k}$ such that

$\displaystyle\frac{1}{h_{k}r_{k}}<\frac{1}{k},$ (4.34a)
$\displaystyle\|\mathrm{sdist}(\cdot,\sigma_{r_{k}}(\partial
A_{m_{h_{k}}}))-\mathrm{sdist}(\cdot,\sigma_{r_{k}}(\partial
A))\|_{L^{\infty}(Q_{4})}<\frac{1}{k},$ (4.34b)
$\displaystyle\int_{Q_{1}\setminus
U_{0}^{\infty}}|\tau(\widetilde{u}_{h_{k}}(r_{k}x))-\tau(\widetilde{u}(r_{k}x))|dx<\frac{1}{k},$
(4.34c)
$\displaystyle\mu_{h_{k}}(\overline{Q_{r_{k}}})\leq\mu_{0}(Q_{r_{k}})+r_{k}^{2}.$
(4.34d)

Notice that by (4.34a), $h_{k}\to\infty$ as $k\to\infty.$

Let

$D_{k}:=\sigma_{r_{k}}(A_{m_{h_{k}}}\cup S)$

and

$w_{k}(x):=\widetilde{u}_{h_{k}}(r_{k}x),\quad x\in Q_{1}.$

Then the number of connected components of $\partial D_{k}$ lying strictly
inside $Q_{4}$ does not exceed $m_{h_{k}},$ and $w_{k}\in GSBD^{2}(Q_{1})$
with $J_{w_{k}}\subset Q_{1}\cap\partial D_{k}.$ By (4.34b) and (4.33a),

$\mathrm{sdist}(\cdot,\partial
D_{k})\to-\mathrm{dist}(\cdot,T_{0})\quad\text{uniformly in $Q_{4}$ as
$k\to\infty.$}$

Moreover, by (4.33b) and (4.34c) $w_{k}\to u_{0}$ a.e. in $Q_{1}\setminus
U_{1}^{\infty}$ and $|w_{k}|\to+\infty$ a.e. in $U_{1}^{\infty}.$ By the
finiteness of

$\frac{d\mu_{0}}{\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits
L}(0)=\lim\limits_{k\to\infty}\frac{\mu_{0}(Q_{r_{k}})}{2r_{k}}$

and (4.34d)

$\mathcal{H}^{1}(Q_{1}\cap\partial
D_{k})=\frac{\mathcal{H}^{1}(Q_{r_{k}}\cap\partial
A_{m_{h_{k}}})}{r_{k}}\leq\frac{\mu_{h_{k}}(\overline{Q_{r_{k}}})}{c_{1}r_{k}}\leq
C:=\frac{2}{c_{1}}\frac{d\mu_{0}}{\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits L}(0)+1$ (4.35)

for all large $k.$ Moreover, by changing variables as $x=r_{k}y$ and using
(4.20) we get

$\int_{Q_{1}}|e(w_{k})|^{2}dx=\int_{Q_{r_{k}}}|e(\widetilde{u}_{k})|^{2}dy\leq
M$

for all $k;$ note that the first equality holds only in dimension two.

Fix $\delta\in(0,1).$ Since $\varphi$ is uniformly continuous, there exists
$k_{\delta}^{0}>0$ such that

$|\varphi(x,\nu)-\varphi(0,\nu)|<\delta,\qquad
x\in\overline{Q_{r_{k}}},\,\,\nu\in\mathbb{S}^{1}.$

Therefore, by the definitions of $D_{k}$ and $\mu_{h},$ the nonnegativity of
$g$ as well as (4.35)

$\displaystyle\frac{\mu_{h_{k}}(\overline{Q_{r_{k}}})}{r_{k}}\geq$
$\displaystyle\int_{Q_{1}\cap\partial^{*}D_{k}}\phi(\nu_{D_{k}})d\mathcal{H}^{1}+2\int_{Q_{1}\cap
D_{k}^{(1)}\cap\partial
D_{k}}\phi(\nu_{D_{k}})d\mathcal{H}^{1}-2Cc_{2}\delta,$ (4.36)

where

$\phi(\nu)=\varphi(0,\nu).$

By Lemma 4.7 applied with sequences $\\{D_{k}\\}$ and $\\{m_{h_{k}}\\}$ we
find $k_{\delta}^{2}>k_{\delta}^{1}$ such that

$\int_{Q_{1}\cap\partial^{*}D_{k}}\phi(\nu_{D_{k}})d\mathcal{H}^{1}+2\int_{Q_{1}\cap
D_{k}^{(1)}\cap\partial D_{k}}\phi(\nu_{D_{k}})d\mathcal{H}^{1}\geq
2\int_{I_{1}}\phi({\bf e_{2}})d\mathcal{H}^{1}-\delta.$

Thus, by (4.36) and (4.34d) we get

$\frac{\mu_{0}(Q_{r_{k}})}{2r_{k}}+\frac{r_{k}}{2}\geq\int_{I_{1}}\phi({\bf
e_{2}})d\mathcal{H}^{1}-\frac{2Cc_{1}+1}{2}\,\delta$

for all $k>k_{\delta}^{2}.$ Now letting first $k\to+\infty$ and then
$\delta\to 0$ we get (4.27). ∎

##  5\. Proof of the main results

The aim of this section is to prove theorems of Section 2.4. We start by
showing that the volume-constraint infima of $\mathcal{F}$ in $\mathcal{C}$
and of $\widetilde{\mathcal{F}}$ in $\widetilde{\mathcal{C}}$ in fact
coincide.

######  Proposition 5.1.

Assume hypotheses (H1)-(H3) and let $\mathtt{v}\in(0,|\Omega|)$ or
$S=\emptyset.$ Then

$\inf\limits_{(A,u)\in\mathcal{C},\,|A|=\mathtt{v}}\,\mathcal{F}(A,u)\,\,=\inf\limits_{(A,u)\in\widetilde{\mathcal{C}},\,|A|=\mathtt{v}}\,\widetilde{\mathcal{F}}(A,u)=\inf\limits_{(A,u)\in\widetilde{\mathcal{C}}}\,\widetilde{\mathcal{F}}^{\lambda}(A,u)$
(5.1)

for any $\lambda\geq\lambda_{0},$ where $\lambda_{0}$ is given by [46, Theorem
2.6] and $\widetilde{\mathcal{F}}^{\lambda}$ is given by (5.17).

###### Proof.

We repeat similar arguments to [46, Section 5]. Note that for any $\lambda>0$

$\inf\limits_{(A,u)\in\mathcal{C},\,|A|=\mathtt{v}}\,\mathcal{F}(A,u)\,\,\geq\inf\limits_{(A,u)\in\widetilde{\mathcal{C}},\,|A|=\mathtt{v}}\,\widetilde{\mathcal{F}}(A,u)\geq\inf\limits_{(A,u)\in\widetilde{\mathcal{C}}}\,\widetilde{\mathcal{F}}^{\lambda}(A,u).$
(5.2)

Further we fix any $\lambda\geq\lambda_{0}.$ Recall that from [46] for such
$\lambda$

$\inf\limits_{(A,u)\in\mathcal{C},\,|A|=\mathtt{v}}\,\mathcal{F}(A,u)=\lim\limits_{m\to+\infty}\,\min\limits_{(A,u)\in\mathcal{C}_{m},\,|A|=\mathtt{v}}\,\mathcal{F}(A,u)=\lim\limits_{m\to+\infty}\,\min\limits_{(A,u)\in\mathcal{C}_{m}}\,\mathcal{F}^{\lambda}(A,u),$

where $\mathcal{F}^{\lambda}$ is given by (2.16). Thus, in view of (5.2) to
prove (5.1) it is enough to establish that $\epsilon>0$ there exists
$n_{\epsilon}\in\mathbb{N}$ and
$(A_{\epsilon},u_{\epsilon})\in\mathcal{C}_{n_{\epsilon}}$ such that

$\inf\limits_{(A,u)\in\widetilde{\mathcal{C}}}\,\widetilde{\mathcal{F}}^{\lambda}(A,u)+\epsilon>\mathcal{F}^{\lambda}(A_{\epsilon},u_{\epsilon}).$
(5.3)

To prove the existence of $n_{\epsilon}$ and
$(A_{\epsilon},u_{\epsilon})\in\mathcal{C}_{n_{\epsilon}}$, we repeat
essentially the same arguments of the proof of [46, Eq. 5.4]. For the
convenience of the reader we give the detailed proof. Given $\epsilon>0$ let
$(B_{1},v_{1})\in\widetilde{\mathcal{C}}$ be such that

$\inf\limits_{(A,u)\in\widetilde{\mathcal{C}}}\,\mathcal{F}^{\lambda}(A,u)>\mathcal{F}^{\lambda}(B_{1},v_{1})-\epsilon.$
(5.4)

Since $|B_{1}|=|\mathrm{Int}(B_{1})|$ and
$\mathcal{F}^{\lambda}(B_{1},v_{1})\geq\mathcal{F}^{\lambda}(\mathrm{Int}(B_{1}),v_{1}),$
we may assume that $B_{1}=\mathrm{Int}(B_{1}),$ i.e., $B_{1}$ is open.

Step 1: First we remove the jump set $J_{v_{1}}$ of $v_{1}$ on $\Sigma$ making
a hole in $\Omega.$ Recall that by our choice, $\nu_{\Sigma}$ is always
directed towards $\Omega.$ Since $\Sigma$ is Lipschitz, by the regularity of
$\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits\Sigma,$ there exists an open
set $\Sigma^{\prime}\subset\Sigma$ such that
$\mathcal{H}^{1}(J_{v_{1}}\setminus\Sigma^{\prime})=0$ and
$\mathcal{H}^{1}(\Sigma^{\prime}\setminus J_{v_{1}})<\frac{\epsilon}{c_{2}}.$

Let $r\in(0,\frac{\epsilon}{\lambda\mathcal{H}^{1}(\Sigma)})$ be such that

$|\varphi(x,\nu)-\varphi(y,\nu)|<\frac{\epsilon}{\mathcal{H}^{1}(\Sigma)}$
(5.5)

whenever $|x-y|<4r.$ Since $\Sigma$ is Lipschitz, by Vitaly Covering Lemma we
can find an at most countable family
$\\{Q_{r_{j},\nu_{\Sigma}(x_{j})}(x_{j})\\}_{j\geq 1}$ of disjoint open
squares such that $x_{j}\in\Sigma,$ $r_{j}\in(0,r),$ $\Sigma\cap
Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})$ is a graph in
$\nu_{\Sigma}(x_{j})$-direction, $\Sigma$ crosses two opposite sides of each
$Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})$ parallel to $\nu_{\Sigma}(x_{j})$ and

$\mathcal{H}^{1}\Big{(}\Sigma^{\prime}\setminus\bigcup_{j}\overline{Q_{r_{j},\nu_{\Sigma}(x_{j})}(x_{j})}\Big{)}=0.$
(5.6)

Note that $\sum\limits_{j}r_{j}<\mathcal{H}^{1}(\Sigma).$ For each $j$ define

$\Sigma_{j}:=(\Sigma\cap\overline{Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})})+\rho_{j}\nu_{\Sigma}(x_{j}),$

where $\rho_{j}\in(0,r_{j})$ is such that $\Sigma_{j}$ still connects two
vertical sides of $Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})$ and
$\sum_{j}\rho_{j}<\frac{\epsilon}{2c_{2}}.$ Let $U_{j}$ be the open set whose
boundaries are $\Sigma_{j},$
$\Sigma\cap\overline{Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})}$ and two vertical sides
of $Q_{r,\nu_{\Sigma}(x_{j})}(x_{j}).$ Note that $\\{U_{j}\\}_{j}$ is a
countable family of pairwise disjoint open sets.

Let $B_{2}:=B_{1}\setminus\overline{\cup_{j}U_{j}}$ and
$v_{2}:=v_{1}\Big{|}_{B_{2}\cup S}.$ Then using the localized version of
$\mathcal{S}$ we get

$\displaystyle\mathcal{S}(B_{2},v_{2})\leq$
$\displaystyle\mathcal{S}(B_{1},v_{1}-u_{0};\Omega\setminus\overline{\cup_{j}U_{j}})+\sum\limits_{j}\Big{(}\int_{\Sigma_{j}}\varphi(x,\nu_{\Sigma}(x))d\mathcal{H}^{1}+2c_{2}\rho_{j}\Big{)}.$
(5.7)

By (5.5) and the definition of $\Sigma_{j}$

$\int_{\Sigma_{j}}\varphi(x,\nu_{\Sigma}(x))d\mathcal{H}^{1}\leq\int_{\Sigma\cap
Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})}\varphi(y,\nu_{\Sigma}(y))d\mathcal{H}^{1}+\frac{\epsilon\mathcal{H}^{1}(\Sigma\cap
Q_{r,\nu_{\Sigma}(x_{j})}(x_{j}))}{\mathcal{H}^{1}(\Sigma)}.$

Thus summing this inequality in $j$ and using pairwise disjointness of
$Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})$ and (5.6) we get

$\sum\limits_{j}\int_{\Sigma_{j}}\varphi(x,\nu_{\Sigma}(x))d\mathcal{H}^{1}\leq\int_{\Sigma^{\prime}}\varphi(y,\nu_{\Sigma}(y))d\mathcal{H}^{1}+\frac{\epsilon\mathcal{H}^{1}(\Sigma^{\prime})}{\mathcal{H}^{1}(\Sigma)}.$

Using the definition of $\Sigma^{\prime}$ we obtain

$\sum\limits_{j}\int_{\Sigma_{j}}\varphi(x,\nu_{\Sigma}(x))d\mathcal{H}^{1}\leq\int_{J_{v_{1}}}\varphi(y,\nu_{\Sigma}(y))d\mathcal{H}^{1}+2\epsilon.$

Inserting this in (5.7) and using the inequality
$\sum_{j}\rho_{j}<\frac{\epsilon}{2c_{2}}$ we get

$\mathcal{S}(B_{2},v_{2})\leq\mathcal{S}(B_{1},v_{1}-u_{0};\Omega\setminus\overline{\cup_{j}U_{j}})+\int_{J_{v_{1}}}\varphi(y,\nu_{\Sigma}(y))d\mathcal{H}^{1}+3\epsilon\leq\mathcal{S}(B_{1},v_{1})+3\epsilon.$

Then by the nonnegativity of the elastic energy, for $(B_{2},v_{2})$ we get

$\widetilde{\mathcal{F}}(B_{2},v_{2})\leq\widetilde{\mathcal{F}}(B_{1},v_{1})+3\epsilon.$

Notice that by our construction $\Sigma\cap J_{v_{2}}$ is
$\mathcal{H}^{1}$-negligible, hence by Lemma A.2 $v_{2}\in
H^{1}_{\mathrm{loc}}({\rm Int}{(B_{2}\cup S\cup\Sigma)}).$

Finally we estimate the volume contribution of $B_{2}.$ Since $U_{j}\subset
Q_{r,\nu_{\Sigma}(x_{j})}(x_{j})$ and $r_{j}\leq
r<\frac{\epsilon}{\lambda\mathcal{H}^{1}(\Sigma)},$ using
$\sum_{j}r_{j}<\mathcal{H}^{1}(\Sigma)$ we get

$|B_{1}\setminus
B_{2}|\leq\sum\limits_{j}|U_{j}|\leq\sum\limits_{j}r_{j}^{2}\leq
r\sum\limits_{j}r_{j}<\frac{\epsilon}{\lambda}.$

Therefore,

$\widetilde{\mathcal{F}}^{\lambda}(B_{1},v_{1})\geq\widetilde{\mathcal{F}}^{\lambda}(B_{2},v_{2})-4\epsilon.$
(5.8)

Step 2: Let $\\{E_{i}\\}_{i\geq 1}$ be all open connected components of
$B_{2}$ (recall that $B_{2}$ is open). We remove all sufficiently small
connected components of $B_{1}.$ Using the localized versions of $\mathcal{S}$
and $\mathcal{W}$ we have

$\mathcal{W}(B_{2},v_{2}-u_{0};\Omega)=\sum\limits_{i\geq
1}\mathcal{W}(E_{i},v_{2}-u_{0};\Omega).$

Since $\partial E_{i}\cap\partial E_{j}\subset B_{2}^{(1)}\cap\partial B_{2}$
and $\varphi(x,\cdot)$ is even,

$\mathcal{S}(B_{2},v_{2};\Omega)=\sum\limits_{i\geq
1}\mathcal{S}(E_{i},v_{2};\Omega).$

Hence, there exists $N_{1}\in\mathbb{N}$ such that the set
$B_{3}:=\cup_{i=1}^{N_{1}}E_{i}$ satisfies

$\begin{gathered}\mathcal{S}(B_{2},v_{2};\Omega)+\mathcal{W}(B_{2},v_{2}-u_{0};\Omega)+\epsilon>\mathcal{S}(B_{3},v_{2};\Omega)+\mathcal{W}(B_{3},v_{2}-u_{0};\Omega),\\\
0\leq|B_{2}|-|B_{3}|<\frac{\epsilon}{\lambda}.\end{gathered}$

Thus,

$\mathcal{F}^{\lambda}(B_{2},v_{2})>\mathcal{F}^{\lambda}(B_{3},v_{3})-2\epsilon,$
(5.9)

where $v_{3}:=v_{2}\Big{|}_{B_{3}}.$

Step 3: Let $\\{F_{j}\\}_{j\geq 1}$ be all connected components of
$\Omega\setminus\overline{B_{3}}$ such that $\partial F_{j}\subset\partial
B_{3}$ (hence, $F_{i}$ are holes in $B_{3}$). We fill in all sufficiently
small holes. Since $\mathcal{S}(B_{3},v)<+\infty,$ there exists $N_{2}\geq 1$
such that

$\sum\limits_{i>N_{2}}\mathcal{S}(F_{i},v_{3};\Omega)+\sum\limits_{i>N_{2}}\mathcal{W}(F_{i},v_{3}-u_{0};\Omega)<\epsilon,\quad\sum\limits_{i>N_{2}}|F_{i}|<\frac{\epsilon}{\lambda}.$

Then the set $B_{4}:=B_{3}\cup(\cup_{i>N_{2}}F_{i})$ and the function
$v_{4}:=v_{3}\chi_{B_{2}\cup S}+u_{0}\chi_{\cup_{i>N_{2}}F_{i}}$ satisfies

$\mathcal{F}^{\lambda}(B_{3},v_{3})>\mathcal{F}^{\lambda}(B_{4},v_{4})-2\epsilon.$
(5.10)

By construction, $\overline{\partial^{*}B_{4}}$ has at most $N_{1}+N_{2}$
connected components.

Step 5: Finally we construct
$(A_{\epsilon},u_{\epsilon})\in\mathcal{C}_{n_{\epsilon}}$ satisfying (5.3)
for some $n_{\epsilon}\in\mathbb{N}.$ Let
$B_{5}:=\mathrm{Int}(\overline{B_{4}}).$ Since $B_{4}$ can have finitely many
“substantial” holes $B_{5}\cap\partial B_{4}=\emptyset.$ In particular, if we
extend $v_{4}$ arbitrarily to the set $B_{4}^{(1)}\cap\partial B_{4}$ and
denote the extension by $v_{5},$ then $v_{5}\in GSBD^{2}({\rm Int}{(B_{5}\cup
S\cup\Sigma)})$ and $J_{v_{5}}=S_{v_{4}}^{B_{4}}$ up to a
$\mathcal{H}^{1}$-negligible set, where $S_{u}^{A}$ is defined in (2.5). Since
$v_{5}=v_{4}$ a.e. in $B_{5},$ by (5.4)-(5.10)

$\int_{B_{5}\cup
S}\mathbb{C}(x)e(v_{5}):e(v_{5})=\mathcal{W}(B_{4},v_{4})\leq\widetilde{\mathcal{F}}(B_{4},v_{4})+c_{2}\mathcal{H}^{1}(\Sigma)\leq
C+9\epsilon,$

where $C:=\max\\{1,\inf_{\widetilde{\mathcal{C}}}\widetilde{\mathcal{F}}\\}$
is independent of $\epsilon.$

By [14, Theorem 1.1] there exists $u_{\epsilon}\in SBV^{2}({\rm
Int}{(B_{5}\cup S\cup\Sigma)})\cap L^{\infty}({\rm Int}{(B_{5}\cup
S\cup\Sigma)})$ such that $J_{u_{\epsilon}}$ is contained in a union $\Gamma$
of finitely many closed connected pieces of $C^{1}$-curves in ${\rm
Int}{(B_{5}\cup S\cup\Sigma)},$ $u_{\epsilon}\in W^{1,\infty}({\rm
Int}{(B_{5}\cup S\cup\Sigma)})$ and

$\int_{B_{5}\cup
S}|e(u_{\epsilon})-e(v_{5})|^{2}dx\leq\frac{\epsilon}{4(C+11\epsilon)(\|\mathbb{C}\|_{\infty}+1)}$
(5.11)

and

$\mathcal{H}^{1}(J_{u_{\epsilon}}\Delta J_{v_{5}})<\frac{\epsilon}{2c_{2}}.$
(5.12)

Since $J_{v_{5}}\subset B_{5},$ we can assume that the squares
$\\{Q_{j}\\}_{j\geq 1}$ of Vitali cover in [14, Eq. 4.3a] satisfies
$Q_{j}\subset\subset B_{5}.$ Therefore, we may assume that
$\Gamma\subset\overline{B_{5}}.$ Since $\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits\Gamma$ is regular, we may
extract finitely many intervals of $\Gamma$ whose union $\Gamma^{\prime}$
still covers $J_{u_{\epsilon}}$ and satisfies
$\mathcal{H}^{1}(\Gamma^{\prime}\setminus
J_{u_{\epsilon}})<\frac{\epsilon}{2c_{2}}.$ Now we define

$A_{\epsilon}:=B_{5}\setminus\overline{\Gamma^{\prime}}.$

Recall that both $\Sigma\cap J_{v_{5}}$ and $\Sigma\cap J_{e-\epsilon}$ are
$\mathcal{H}^{1}$-negligible. By the definition of $B_{5}$ and
$\Gamma^{\prime},$ there exists $n_{\epsilon}\in\mathbb{N}$ such that
$(A_{\epsilon},u_{\epsilon})\in\mathcal{C}_{n_{\epsilon}}.$ By the definition
of $\widetilde{\mathcal{S}},$ $B_{5}$ and $v_{5}$ as well as by (5.12) we have

$\displaystyle\widetilde{\mathcal{S}}(B_{4},v_{4})=$
$\displaystyle\int_{\Omega\cap\partial^{*}B_{5}}\varphi(x,\nu_{B_{5}})d\mathcal{H}^{1}+2\int_{B_{5}\cap
J_{v_{5}}}\varphi(x,\nu_{J_{v_{5}}})d\mathcal{H}^{1}+\int_{\Sigma\cap\partial^{*}B_{5}}\beta
d\mathcal{H}^{1}$ $\displaystyle\geq$
$\displaystyle\int_{\Omega\cap\partial^{*}B_{5}}\varphi(x,\nu_{B_{5}})d\mathcal{H}^{1}+2\int_{B_{5}\cap
J_{u_{\epsilon}}}\varphi(x,\nu_{J_{u_{\epsilon}}})d\mathcal{H}^{1}+\int_{\Sigma\cap\partial^{*}B_{5}}\beta
d\mathcal{H}^{1}-\epsilon.$

Thus, by the definition of $A_{\epsilon}$ and $\Gamma^{\prime}$

$\displaystyle\widetilde{\mathcal{S}}(B_{4},v_{4})\geq$
$\displaystyle\int_{\Omega\cap\partial^{*}A_{\epsilon}}\varphi(x,\nu_{A_{\epsilon}})d\mathcal{H}^{1}+2\int_{A_{\epsilon}^{(1)}\cap\Gamma^{\prime}}\varphi(x,\nu_{\Gamma^{\prime}})d\mathcal{H}^{1}+\int_{\Sigma\cap\partial^{*}A_{\epsilon}}\beta
d\mathcal{H}^{1}-2\epsilon$ $\displaystyle=$
$\displaystyle\mathcal{S}(A_{\epsilon},u_{\epsilon})-2\epsilon.$ (5.13)

Moreover, using the relations $|A_{\epsilon}\Delta B_{4}|=0$ and $v_{4}=v_{5}$
a.e. in $B_{5}$ and Cauchy-Schwartz inequality for nonnegative symmetric forms
we obtain

$\displaystyle\mathcal{W}(A_{\epsilon},u_{\epsilon})\leq$
$\displaystyle\mathcal{W}(B_{4},v_{4})+2\int_{B_{5}\cup
S}\mathbb{C}(x)e(u_{\epsilon}):(e(u_{\epsilon})-e(v_{5}))$ $\displaystyle\leq$
$\displaystyle\mathcal{W}(B_{4},v_{4})+2\Big{(}\int_{B_{5}\cup
S}\mathbb{C}(x)e(u_{\epsilon}):e(u_{\epsilon})dx\Big{)}^{1/2}\times$
$\displaystyle\times\Big{(}\int_{B_{5}\cup
S}\mathbb{C}(x)(e(u_{\epsilon})-e(v_{5})):(e(u_{\epsilon})-e(v_{5}))dx\Big{)}^{1/2}.$
(5.14)

Similarly,

$\displaystyle\int_{B_{4}\cup
S}\mathbb{C}(x)e(u_{\epsilon}):e(u_{\epsilon})dx$ $\displaystyle\leq$
$\displaystyle\mathcal{W}(B_{4},v_{4})+2\Big{(}\mathcal{W}(B_{4},v_{4})\Big{)}^{1/2}\Big{(}\int_{B_{5}\cup
S}\mathbb{C}(x)(e(u_{\epsilon})-e(v_{5})):(e(u_{\epsilon})-e(v_{5}))dx\Big{)}^{1/2}$
$\displaystyle\leq$
$\displaystyle(C+9\epsilon)+2\sqrt{(C+9\epsilon)\|\mathbb{C}\|_{\infty}}\,\,\|e(u_{\epsilon})-e(v_{5})\|_{L^{2}}\leq
C+10\epsilon,$

where in the last inequality we used (5.11). Therefore, by (5) and again by
(5.11)

$\displaystyle\mathcal{W}(A_{\epsilon},u_{\epsilon})\leq$
$\displaystyle\mathcal{W}(B_{4},v_{4})+2\sqrt{(C+10\epsilon)\|\mathbb{C}\|_{\infty}}\,\,\|e(u_{\epsilon})-e(v_{5})\|_{L^{2}}\leq\mathcal{W}(B_{4},v_{4})+\epsilon.$
(5.15)

Now combining (5) and (5.15) as well as using $|B_{5}|=|A_{\epsilon}|$ we get

$\widetilde{\mathcal{F}}^{\lambda}(B_{4},v_{4})\geq\mathcal{F}^{\lambda}(A_{\epsilon},u_{\epsilon})-3\epsilon.$
(5.16)

Since $(A_{\epsilon},u_{\epsilon})\in\mathcal{C}_{n_{\epsilon}},$ by (5.4),
(5.8), (5.9), (5.10) and (5.16) we get

$\inf\limits_{(A,u)\in\widetilde{\mathcal{C}}}\,\widetilde{\mathcal{F}}(A,u)+12\epsilon\geq\mathcal{F}(A_{\epsilon},u_{\epsilon}),$

and (5.3) follows. ∎

Proposition 5.1 implies that the configuration $(A,u)$ given by Proposition
4.6 is a volume-constraint minimizer of $\widetilde{\mathcal{F}}$ in
$\widetilde{\mathcal{C}}.$

######  Proposition 5.2.

Let $(A,u)\in\widetilde{\mathcal{C}}$ be given by Proposition 4.6. Then
$(\mathrm{Int}(A),u)$ is a minimizer of $\widetilde{\mathcal{F}}$ in
$\widetilde{\mathcal{C}}$ under the volume constraint $|A|=\mathtt{v}.$
Moreover, let $\lambda_{0}$ be as in Proposition 5.1 and let
$(B,v)\in\widetilde{\mathcal{C}}$ be any volume-constraint minimizer of
$\widetilde{\mathcal{F}}.$ Then $(B,v)$ is a minimizer of
$\widetilde{\mathcal{F}}^{\lambda}$ for all $\lambda\geq\lambda_{0},$ where

$\widetilde{\mathcal{F}}^{\lambda}(B,v):=\widetilde{\mathcal{F}}(B,v)+\lambda\big{|}|B|-\mathtt{v}\big{|},\quad(B,v)\in\widetilde{\mathcal{C}},\quad\lambda>0.$
(5.17)

###### Proof.

Note that since $|\mathrm{Int}(A)\Delta A|=0$ and
$(\mathrm{Int}(A),u)\in\widetilde{\mathcal{C}},$ by Propositions 4.5, 4.6 and
5.1

$\widetilde{\mathcal{F}}(\mathrm{Int}(A),u)=\inf\limits_{(B,v)\in\mathcal{C}}\,\mathcal{F}(B,v)=\inf\limits_{(B,v)\in\widetilde{\mathcal{C}}}\widetilde{\mathcal{F}}(B,v)=\inf\limits_{(B,v)\in\mathcal{C}}\widetilde{\mathcal{F}}^{\lambda}(B,v)$

for all $\lambda\geq\lambda_{0}.$ Thus, $(\mathrm{Int}(A),u)$ is a minimizer
of both $\widetilde{\mathcal{F}}$ and $\widetilde{\mathcal{F}}^{\lambda_{0}}.$
The same is true for every minimizer $(B,v)$ of $\widetilde{\mathcal{F}}.$ ∎

######  Theorem 5.3 (Density estimates for minimizers of
$\widetilde{\mathcal{F}}^{\lambda}$).

Given $\lambda>0,$ let $(A,u)\in\widetilde{\mathcal{C}}$ be any minimizer of
$\widetilde{\mathcal{F}}^{\lambda}(\cdot,\cdot)$ in $\widetilde{\mathcal{C}}$
and let $\xi\in\mathbb{R}^{2}$ be such that for the function

$\widetilde{u}:=u\chi_{A\cup S}+\xi\chi_{\Omega\setminus A}$

one has $\Omega\cap\partial^{*}A\subset J_{\widetilde{u}}.$ Then for any
$x\in\Omega$ and $r\in(0,\mathrm{dist}(x,\partial\Omega)),$

$\frac{\mathcal{H}^{1}(Q_{r}(x)\cap
J_{\widetilde{u}})}{r}\leq\frac{16c_{2}+4\lambda}{c_{1}}.$ (5.18)

Moreover, there exist $\varsigma^{*}=\varsigma^{*}\in(0,1)$ and $R^{*}>0$ not
depending on $(A,u)$ with the following property. If $x\in\Omega$ belongs to
the closure $J^{c}_{\widetilde{u}}$ of the set $\\{y\in\Omega\cap
J_{\widetilde{u}}:\,\theta_{*}(J_{\widetilde{u}},y)>0\\},$ then

$\frac{\mathcal{H}^{1}(Q_{r}(x)\cap J_{\widetilde{u}})}{r}\geq\varsigma^{*}$
(5.19)

for any square $Q_{r}(x)\subset\subset\Omega$ with
$r\in(0,\min\\{R^{*},\mathrm{dist}(x,\partial\Omega)\\}),$ and if $x\in\Omega$
belongs to the closure $S_{u}^{c}$ of $\\{x\in
S_{u}^{A}:\,\theta_{*}(S_{u}^{A},x)>0\\},$ then

$\frac{\mathcal{H}^{1}(Q_{r}(x)\cap S_{u}^{A})}{r}\geq\varsigma^{*}$ (5.20)

for any $r\in(0,\min\\{R^{*},\mathrm{dist}(x,\overline{\partial^{*}A})\\}.$ In
particular,

$\mathcal{H}^{1}(\Omega\cap(J_{\widetilde{u}}^{c}\setminus
J_{\widetilde{u}})=\mathcal{H}^{1}(\mathrm{Int}(A^{(1)})\cap(S_{u}^{c}\setminus
S_{u}^{A})=0.$ (5.21)

###### Proof of Theorem 5.3.

As in Remark 2.5 $(A,u)$ is a minimizer of $\widetilde{\mathcal{F}}^{\lambda}$
if and only if $(A,u+u_{0})$ minimizes the
$\widehat{\widetilde{\mathcal{F}}^{\lambda}}(\cdot):=\widetilde{\mathcal{F}}^{\lambda}(\cdot-
u_{0}).$ Thus, we can introduce the following localized version of
$\widetilde{\mathcal{F}}$ in open subsets $O$ of $\Omega$ which does not see
the substrate:

$\widetilde{\mathcal{F}}(B,v;O):=\widetilde{\mathcal{S}}(B;O)+\mathcal{W}(B,v;O)$

where

$\widetilde{\mathcal{S}}(B,v;O):=\int_{O\cap\partial^{*}B}\varphi(y,\nu_{B})d\mathcal{H}^{1}+2\int_{O\cap
B^{(1)}\cap\partial B\cap S_{v}}\varphi(y,\nu_{B})d\mathcal{H}^{1},$

the $\mathcal{W}(\cdot;O)$ is given as in (2.8) and $S_{v}^{A}$ is defined as
in (2.5). Then the minimality of $(A,u)$ implies that $(A,u+u_{0})$ is a
quasi-minimizer of $\widetilde{\mathcal{F}}(\cdot;O)$ in $O,$ namely,

$\widetilde{\mathcal{F}}(A,u+u_{0};O)\leq\widetilde{\mathcal{F}}(B,v;O)+\lambda_{0}|A\Delta
B|$

whenever $(B,v)\in\widetilde{\mathcal{C}}$ with $A\Delta B\subset\subset O$
and $\mathrm{supp}\,(u+u_{0}-v)\subset\subset O.$ Now the proof of the
existence of $\varsigma^{*}$ and $R^{*}$ satisfying (5.18) and (5.19) runs
along the same lines of the proof of Theorem 3.1 for $m=\infty,$ therefore, we
do not repeat it here. Note that $\varsigma^{*}$ and $R^{*}$ depend only on
$c_{i}$ and $\lambda.$

Let $A_{\circ}:=\mathrm{Int}(A^{(1)}).$ We claim that

$\partial A_{\circ}=\overline{\partial^{*}A}.$

Indeed, note that $A^{(1)}\setminus A_{\circ}\subset\partial
A^{(1)}=\overline{\partial^{*}A},$ where in the equality we used
$\overline{\partial^{*}A}=\overline{\partial^{*}A^{(1)}}=\partial A^{(1)}$ see
e.g., [51, Eq. 15.3]. Thus, $A_{\circ}$ is also equivalent to $A,$ and hence,
$\partial^{*}A_{\circ}=\partial^{*}A=\partial^{*}A^{(1)}.$ In particular,
$\partial A^{(1)}=\overline{\partial^{*}A_{\circ}}\subset\partial A_{\circ}.$
On the other hand, assume that there exists $x\in\partial
A_{\circ}\setminus\partial A^{(1)}.$ Since $\partial A^{(1)}$ is closed, there
exists $r>0$ such that $\overline{B_{r}(x)}\cap\partial A^{(1)}=\emptyset.$
Hence, either $B_{r}(x)\subset\mathrm{Int}(A^{(1)})=A_{\circ}$ or
$\overline{B_{r}(x)}\cap\overline{A^{(1)}}=\emptyset.$ Since $A_{\circ}$ is
open and $x\in\partial A_{\circ},$ the inclusion $B_{r}(x)\subset A_{\circ}$
is not possible. On the other hand, since
$\overline{A_{\circ}}\subset\overline{A^{(1)}}$ and $x\in\partial A_{\circ},$
the relation $\overline{B_{r}(x)}\cap\overline{A^{(1)}}=\emptyset$ is also not
possible. Thus, $\partial A_{\circ}\subseteq\partial A^{(1)}.$

To prove (5.20) we fix $\Omega^{\prime}\subset\subset\Omega.$ We claim that
$\widetilde{u}\big{|}_{A_{\circ}}$ is a minimizer of Griffith functional
$\mathcal{G}:GSBD^{2}({\rm Int}{(A_{\circ}\cup S\cup\Sigma)})\to\mathbb{R},$

$\mathcal{G}(v):=\int_{A_{\circ}\cap
J_{v}}\varphi(x,\nu_{J_{v}})d\mathcal{H}^{1}+\int_{A_{\circ}}\mathbb{C}(x)e(v):e(v)dx$

with Dirichlet boundary condition $v=\widetilde{u}=u$ in
$A_{\circ}\setminus\Omega^{\prime}.$ Indeed, for every $v\in
GSBD^{2}(A_{\circ})$ with $\widetilde{u}=v$ in
$A_{\circ}\setminus\Omega^{\prime}$ we define
$B:=A_{\circ}\setminus\overline{J_{v}}.$ Then
$(B,v)\in\widetilde{\mathcal{C}}$ and by the minimality of $(A,u)$

$\mathcal{G}(u)-\mathcal{G}(v)=\widetilde{\mathcal{F}}(A,u)-\widetilde{\mathcal{F}}(B,v)\leq
0.$

Since $S_{v}^{B}=J_{\widetilde{u}\big{|}_{A_{\circ}}}$ up to a
$\mathcal{H}^{1}$-negligible set, (5.20) directly follows from the density
estimates for the jump set of Griffith minimizers (see e.g. [13]) with
possibly smaller $\varsigma^{*}\in(0,1)$ and $R^{*}>0.$

Finally, we prove (5.21) only for $S_{u}^{A},$ the other being similar. Let
$\Gamma:=\\{x\in S_{u}^{A}:\,\theta_{*}(S_{u}^{A},x)>0\\}.$ Note that
$S_{u}^{c}=\bar{\Gamma}.$

We claim that

$\mathcal{H}^{1}(A_{\circ}\cap(\bar{\Gamma}\setminus\Gamma))=0.$ (5.22)

Indeed, let $\mu:=\mathcal{H}^{1}\mathop{\hbox{\vrule
height=9.0pt,width=0.5pt,depth=0.0pt\vrule
height=0.5pt,width=5.0pt,depth=0.0pt}}\nolimits\Gamma.$ Then
$\mu(\bar{\Gamma}\setminus\Gamma)=0.$ By the regularity of $\mu,$ for every
$\epsilon>0$ there exists an open set $U\subset\mathbb{R}^{2}$ such that
$L:=A_{\circ}\cap(\bar{\Gamma}\setminus\Gamma)\subset U$ and
$\mu(U)=\mathcal{H}^{1}(U\cap\Gamma)<\epsilon.$ Note that
$\bar{\Gamma}\subset\overline{\\{y\in\Omega\cap
J_{\widetilde{u}}:\,\theta_{*}(J_{\widetilde{u}},x)>0\\}},$ where
$\widetilde{u}$ is given by Theorem 5.3. Hence, for (5.19) holds for all
points of $\bar{\Gamma}.$ By the definition of the closure, and Vitali
Covering Lemma we can find at most countable pairwise disjoint family
$\\{\overline{B_{r_{i}}(x_{i})}\\}_{i}$ of closed balls
$\overline{B_{r_{i}}(x_{i})}$ with $x_{i}\in A_{\circ}\cap\Gamma,$
$r_{i}\leq\min\\{R^{*},\epsilon,\mathrm{dist}(x,\partial\bar{A})\\}$ such that
$A_{\circ}\cap(\bar{\Gamma}\setminus\Gamma)\subset\cup_{i}\overline{B_{5r_{i}}(x_{i})}.$
Without loss of generality we may assume that $B_{r_{i}}(x_{i})\subset U.$
Since $Q_{r_{i}/\sqrt{2}}(x_{i})\subset B_{r_{i}}(x_{i})\subset
Q_{r_{i}}(x_{i}),$ from the definition of Hausdorff premeasure, (5.19) and
disjointness of $\\{B_{r_{i}}(x_{i})\\}$ as well as the choice of $U$ we
obtain

$\displaystyle\mathcal{H}_{10\epsilon}(A_{\circ}\cap(\bar{\Gamma}\setminus\Gamma))\leq$
$\displaystyle\sum\limits_{i\geq
1}2\pi(5r_{i})\leq\frac{10\pi\sqrt{2}}{\varsigma^{*}}\sum\limits_{i\leq
1}\mathcal{H}^{1}(Q_{r_{i}/\sqrt{2}}(x_{i})\cap\Gamma)$ $\displaystyle=$
$\displaystyle\frac{10\pi\sqrt{2}}{\varsigma^{*}}\,\mathcal{H}^{1}\big{(}\cup_{i}Q_{r_{i}/\sqrt{2}}(x_{i})\cap\Gamma\big{)}\leq\frac{10\pi\sqrt{2}}{\varsigma^{*}}\,\mathcal{H}^{1}\big{(}U\cap\Gamma\big{)}<\frac{10\pi\sqrt{2}\epsilon}{\varsigma^{*}}.$

Now letting $\epsilon\to 0$ we get (5.22). ∎

In the following proposition we construct a “regular” minimizer of
$\mathcal{F}$ starting from a minimizer of $\widetilde{\mathcal{F}}$ in
$\widetilde{\mathcal{C}}.$

######  Proposition 5.4.

Given $\lambda>0,$ let $(A,u)\in\widetilde{\mathcal{C}}$ be any minimizer of
$\widetilde{\mathcal{F}}^{\lambda}.$ Then there exists
$(A^{\prime},u^{\prime})\in\mathcal{C}$ such that

$\widetilde{\mathcal{F}}(A,u)=\mathcal{F}(A^{\prime},u^{\prime}).$

Moreover:

  * (1)

$A^{\prime}$ is open, $\theta_{*}(S_{u^{\prime}}^{A^{\prime}},x)>0$ for all
$x\in S_{u^{\prime}}^{A^{\prime}},$ $|A\Delta A^{\prime}|=0$ and $u\chi_{A\cup
S}=u^{\prime}\chi_{A^{\prime}\cup S}$ a.e. in $\Omega\cup S;$

  * (2)

the closure of $A^{\prime}\cap\partial A^{\prime}$ coincide with
$\overline{S_{u^{\prime}}^{A^{\prime}}}$ and
$\mathcal{H}^{1}(\overline{S_{u^{\prime}}^{A^{\prime}}}\setminus
S_{u^{\prime}}^{A^{\prime}})=0;$

  * (3)

let $\varsigma*$ and $R^{*}$ be given by Theorem 5.3. Then

$\frac{\mathcal{H}^{1}(Q_{r}(x)\cap\partial
A^{\prime})}{r}\leq\frac{16c_{2}+4\lambda_{0}}{c_{1}}$ (5.23)

for every square $Q_{r}(x)\subset\Omega$ and

$\frac{\mathcal{H}^{1}(Q_{r}(x)\cap\partial A)}{r}\geq\varsigma*$ (5.24)

for every $Q_{r}(x)\subset\Omega$ with for any $x\in\partial A^{\prime}$ and
$r\in(0,R^{*}).$

###### Proof.

Let $\Gamma:=\\{x\in S_{u}^{A}:\,\theta_{*}(S_{u}^{A},x)>0\\}$ and let

$A^{\prime}:=\mathrm{Int}(A^{(1)})\setminus\bar{\Gamma}$

and let $u^{\prime}$ be the extension of $u$ to $A^{\prime}\cap\partial A$
such that $J_{u^{\prime}}=S_{u}^{A}.$ Note that $A^{\prime}$ is open,
$|A^{\prime}\Delta A|=0$ and $u\chi_{A\cup S}=u^{\prime}\chi_{A^{\prime}\cup
S}$ a.e. in $\Omega\cup S.$ Moreover,
$S_{u^{\prime}}^{A^{\prime}}\subset\Gamma,$ and by (5.20) all points of
$\Omega\cap\bar{\Gamma}$ satisfy uniform lower density estimates, hence,
$\theta_{*}(S_{u^{\prime}}^{A^{\prime}},x)>0$ for any $x\in
S_{u^{\prime}}^{A^{\prime}}.$

We claim that $A^{\prime}\in\mathcal{A}.$ Indeed, let $\widetilde{u}$ be given
as in Theorem 5.3. By definition

$\Omega\cap J_{\widetilde{u}}^{c}=\Omega\cap\partial
A^{\prime}\quad\text{and}\quad\partial A^{\prime}\subset
J_{\widetilde{u}}^{c}\cup\Sigma,$ (5.25)

where $J_{\widetilde{u}}^{c}$ is the closure of the set $\\{x\in
J_{\widetilde{u}}:\,\theta_{*}(J_{\widetilde{u}},x)>0\\}.$ Since
$J_{\widetilde{u}}$ is $\mathcal{H}^{1}$-rectifiable, so is
$J_{\widetilde{u}}^{c}$ in view of (5.21). Therefore, $\partial A^{\prime}$ is
$\mathcal{H}^{1}$-rectifiable, i.e., $A^{\prime}\in\mathcal{A}.$ Note that by
construction $\mathcal{H}^{1}(A^{\prime}\cap J_{\widetilde{u}})=0$ hence, by
Lemma A.2 $\widetilde{u}\in H_{\mathrm{loc}}^{1}(A^{\prime})$ and, since
$u=\widetilde{u}$ a.e. in $A^{\prime}$ it follows that $u^{\prime}\in
H_{\mathrm{loc}}^{1}(A^{\prime}).$

Since $|A\Delta A^{\prime}|=0$ and $u^{\prime}=u$ a.e. in $A^{\prime},$ it
follows that

$\mathcal{W}(A,u)=\mathcal{W}(A^{\prime},u^{\prime}).$

Moreover, by the definition of $\Gamma$ and $S_{u}^{A},$

$|\mathcal{S}(A^{\prime},u^{\prime})-\widetilde{\mathcal{S}}(A,u)|=\int_{\mathrm{Int}(A^{(1)})\cap(\bar{\Gamma}\setminus
S_{u}^{A})}\varphi(x,\nu_{\Gamma})d\mathcal{H}^{1}\leq
c_{2}\mathcal{H}^{1}(\mathrm{Int}(A^{(1)})\cap(\bar{\Gamma}\setminus\Gamma))=0,$

where in the last equality we used (5.21). Finally, (5.23) and (5.24) follows
from (5.25) and density estimates of Theorem 5.3.

Hence, $(A^{\prime},u^{\prime})$ satisfies all properties stated in the
proposition. ∎

Now we are ready to prove the existence of global minimizers of $\mathcal{F}.$

###### Proof of Theorem 2.6.

First we prove the assertion for $\mathcal{G}=\mathcal{F}.$

Let $(A_{m},u_{m})\in\mathcal{C}_{m}$ be a minimizer of $\mathcal{F}$
satisfying the volume constraint $|A_{m}|=\mathtt{v}$ and let
$(A_{m_{h}},u_{m_{h}}),$ and $A$ and $u$ be as in Proposition 4.6. By (3.1),
(4.13) and (4.12) we have

$\inf\limits_{(B,v)\in\mathcal{C},\,|B|=\mathtt{v}}\mathcal{F}(B,v)=\lim\limits_{h\to+\infty}\mathcal{F}(A_{m_{h}},u_{m_{h}})\geq\widetilde{\mathcal{F}}(\mathrm{Int}(A),u).$

Since $|\mathrm{Int}(A)|=\mathtt{v},$ by Propositions 5.1 and 5.2

$\inf\limits_{(B,v)\in\mathcal{C},\,|B|=\mathtt{v}}\mathcal{F}(B,v)=\inf\limits_{(B,v)\in\widetilde{\mathcal{C}},\,|B|=\mathtt{v}}\widetilde{\mathcal{F}}^{\lambda_{0}}(B,v)=\widetilde{\mathcal{F}}^{\lambda_{0}}(\mathrm{Int}(A),u)=\widetilde{\mathcal{F}}(\mathrm{Int}(A),u),$
(5.26)

hence, $(\mathrm{Int}(A),u)$ is a minimizer of
$\widetilde{\mathcal{F}}^{\lambda_{0}}$ in $\widetilde{\mathcal{C}}.$ Then by
Proposition 5.4 there exists $(A^{\prime},u^{\prime})\in\mathcal{C}$ such that

$\widetilde{\mathcal{F}}(\mathrm{Int}(A),u)=\mathcal{F}(A^{\prime},u^{\prime}),$

and hence, in view of (5.26), $(A^{\prime},u^{\prime})$ is a solution to
(2.15).

The proof of the second assertion (i.e., the existence of $\lambda_{1}$ for
which the set of minimizers in $\mathcal{C}$ of both $\mathcal{F}$ and
$\mathcal{F}^{\lambda}$ coincide for all $\lambda\geq\lambda_{1}$) can be done
using the first one and also following the arguments of [33, Theorem 1.1] and
[46, Proposition A.6]. Without loss of generality we assume that
$\lambda_{1}\geq\lambda_{0},$ where $\lambda_{0}$ is given by Proposition 5.1.

Now we prove Theorem 2.6 for $\mathcal{G}=\widetilde{\mathcal{F}}.$ We have
already shown above that the configuration $(\mathrm{Int}(A),u)$ given by
Proposition 4.6 solves the minimum problem (2.15) with
$\mathcal{G}=\widetilde{\mathcal{F}}.$ In view of (5.1) every volume-
constraint minimizer of $\widetilde{\mathcal{F}}$ also minimizer of
$\widetilde{\mathcal{F}}^{\lambda}$ for all $\lambda\geq\lambda_{1}.$ To prove
the converse assertion, we fix any minimizer $(A,u)\in\widetilde{\mathcal{C}}$
of $\widetilde{\mathcal{F}}^{\lambda}$ for $\lambda\geq\lambda_{1}.$ By
Proposition 5.4 there exists $(A^{\prime},u^{\prime})\in\mathcal{C}$ such that
$|A^{\prime}|=|A|$ and
$\mathcal{F}(A^{\prime},u^{\prime})=\widetilde{\mathcal{F}}(A,u).$ By the
first part of the proof and (5.1) we know that

$\inf\limits_{(B,v)\in\mathcal{C}}\mathcal{F}^{\lambda}(B,v)=\inf\limits_{(B,v)\in\mathcal{C},|B|=\mathtt{v}}\mathcal{F}(B,v)=\inf\limits_{(B,v)\in\widetilde{\mathcal{C}}}\widetilde{\mathcal{F}}^{\lambda}(B,v)=\mathcal{F}^{\lambda}(A^{\prime},u^{\prime}).$

Hence, $(A^{\prime},u^{\prime})$ is the minimizer of $\mathcal{F}^{\lambda}.$
Since $\lambda\geq\lambda_{1}$ according to the first part of the proof,
$|A^{\prime}|=\mathtt{v}.$ Hence, $|A|=\mathtt{v}$ and $(A,u)$ minimizer of
(2.15). ∎

We are ready now to study the properties of the minimizers of $\mathcal{F}$ in
$\mathcal{C}$ provided by Theorem 2.6.

###### Proof of Theorem 2.7.

First we properties (1)-(4) the assertion for
$\mathcal{G}=\widetilde{\mathcal{F}}.$

Consider any solution $(A,u)\in\widetilde{\mathcal{C}}$ of (2.15). By
Proposition 5.4 there exists a $(A^{\prime},u^{\prime})\in\mathcal{C}$ with
$A^{\prime}$ defined as in (2.17), such that the properties (1)-(4) hold
except the conditions $\mathcal{H}^{1}(\partial A\Delta\partial A^{\prime})=0$
and $\mathcal{H}^{1}(S_{u}^{A}\Delta S_{u^{\prime}}^{A^{\prime}})=0$ of (1).
To prove these two equations it is enough to observe that

$0=|\mathcal{F}(A,u)-\mathcal{F}(A^{\prime},u^{\prime})|=2\int_{A^{(1)}\cap(\partial
A\Delta\partial A^{\prime})}\varphi(x,\nu_{A})d\mathcal{H}^{1}$

and

$0=|\widetilde{\mathcal{F}}(A,u)-\widetilde{\mathcal{F}}(A^{\prime},u^{\prime})|=2\int_{A^{(1)}\cap(S_{u}^{A}\Delta
S_{u^{\prime}}^{A^{\prime}})}\varphi(x,\nu_{A})d\mathcal{H}^{1}.$

Now we assume that $\mathcal{G}=\mathcal{F}$ and let $(A,u)\in\mathcal{C}$ be
a solution to (2.15). Since $(A,u)\in\widetilde{\mathcal{C}},$ by Proposition
5.1

$\inf\limits_{(B,v)\in\widetilde{\mathcal{C}},|B|=\mathtt{v}}\widetilde{\mathcal{F}}(B,v)=\mathcal{F}(A,u)\geq\widetilde{\mathcal{F}}(A,u).$

Therefore, $(A,u)$ is also a volume-constraint minimizer of
$\widetilde{\mathcal{F}}.$ Thus, applying first part of the proof we establish
that $(A^{\prime},u^{\prime})\in\mathcal{C}$ satisfies (1)-(4).

Finally, notice that if $E\subset A^{\prime}$ is a connected component of
$A^{\prime}$ with $\mathcal{H}^{1}(\partial E\cap\Sigma\setminus J_{u})=0,$
then for $(A^{\prime},v)$ with $v=u^{\prime}\chi_{(A\cup S)\setminus
E}+(u_{0}+a)\chi_{E},$ where $a$ is any rigid displacement, we have

$\mathcal{S}(A^{\prime},u^{\prime})\geq\mathcal{S}(A^{\prime},v)$

and

$\mathcal{W}(A^{\prime},u^{\prime})\geq\mathcal{W}(A^{\prime},v),$ (5.27)

where in (5.27) equality holds if and only of $u^{\prime}=u_{0}+a$ in $E.$
Therefore, by the minimality of $(A^{\prime},u^{\prime})$ it follows that
$u^{\prime}=u_{0}+a$ in $E.$ It remains to prove

$|E|\geq 4\pi\,\Big{(}\frac{c_{1}}{\lambda_{0}}\Big{)}^{2}.$ (5.28)

Consider the competitor $(A^{\prime}\setminus E,u^{\prime})\in\mathcal{C}.$ By
minimality and Theorem 2.6,
$\mathcal{F}^{\lambda_{1}}(A^{\prime},u^{\prime})\leq\mathcal{F}^{\lambda_{1}}(A^{\prime}\setminus
E,u^{\prime}),$ so that by (5.27) and the additivity of the surface energy,
$\mathcal{S}(E,u^{\prime})\leq\lambda_{1}|E|.$ Then by (2.12) and the
isoperimetric inequality in $\mathbb{R}^{2}$

$\lambda_{1}|E|\geq c_{1}\mathcal{H}^{1}(\partial E)\geq
c_{1}\sqrt{4\pi}|E|^{1/2}.$

Hence, (5.28) follows. ∎

## Appendix A

######  Lemma A.1.

Let $K\subset\mathbb{R}^{2}$ be such that $\mathcal{H}^{1}(K)<+\infty$ and let
$\\{E_{t}\\}_{t\in\varUpsilon}$ be a family of sets parametrized by
$t\in\varUpsilon$ such that

$\mathcal{H}^{1}(K\cap E_{t}\cap E_{s})=0$ (A.1)

and $\mathcal{H}^{1}(K\cap E_{t})>0.$ Then $\varUpsilon$ is at most countable.

###### Proof.

The proof runs along the lines of the proof of [51, Proposition 2.16]. For
$j\in\mathbb{N}$ let $\varUpsilon_{j}\subset\varUpsilon$ be the set of all
$t\in\varUpsilon$ such that $\mathcal{H}^{1}(K\cap E_{t})>\frac{1}{j}.$ Then
by (A.1) $\varUpsilon_{j}$ cannot contain more than $j\mathcal{H}^{1}(K)$
elements. Since $\varUpsilon=\cup_{j}\varUpsilon_{j},$ the set $\varUpsilon$
is at most countable. ∎

######  Lemma A.2.

Let $U\subset\mathbb{R}^{n}$ be a connected bounded open set and $u\in
GSBD^{2}(U)$ be such that $\mathcal{H}^{n-1}(J_{u})=0.$ Then $u\in
H_{\mathrm{loc}}^{1}(U).$

###### Proof.

Indeed, for $r>0$ let $Q:=x_{0}+(-r,r)^{n}\subset U$ be any cube centered at
$x\in U$ and let $0<\theta^{\prime\prime}<\theta^{\prime}<1.$ For shortness,
write $Q^{\prime}:=x_{0}+(-\theta^{\prime}r,\theta^{\prime}r)^{n}$ and
$Q^{\prime\prime}:=x_{0}+(-\theta^{\prime\prime}r,\theta^{\prime\prime}r)^{n}.$
By [12, Proposition 3.1 (1)] (see also [11, Theorem 1.1]) there exists a
$\mathcal{L}^{2}$-measurable set $\omega\subset Q^{\prime}$ and a rigid
displacement $a:\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that $|\omega|\leq
c_{*}r\mathcal{H}^{n-1}(J_{u})=0$ and

$\int_{Q^{\prime}}|u-a|^{\frac{2n}{n-1}}dx=\int_{Q^{\prime}\setminus\omega}|u-a|^{\frac{2n}{n-1}}dx\leq
c_{*}r^{2}\Big{(}\int_{Q}|e(u)|^{2}\Big{)}^{\frac{n}{n-1}},$

where $c_{*}$ depends only on $n.$ Hence, $u\in
L_{\mathrm{loc}}^{\frac{2n}{n-1}}(Q).$ Next, fix any mollifier $\rho_{1}\in
C^{\infty}(B_{r}(0))$ with $\rho_{\epsilon}\in
C_{c}^{\infty}(B_{(\theta^{\prime\prime}-\theta^{\prime})\epsilon}),$ where
$\rho_{\epsilon}(x):=\rho_{1}(x/\epsilon),$ $\epsilon\in(0,r).$ By [12,
Proposition 3.1 (2)] there exists $\bar{p}>0$ depending on $n$ and $\epsilon$
such that

$\int_{Q^{\prime\prime}}|e(u*\rho_{\epsilon})-e(u)*\rho_{\epsilon}|^{2}dx\leq
c\Big{(}\frac{\mathcal{H}^{n-1}(J_{u})}{r^{n-1}}\Big{)}^{\bar{p}}\int_{Q}|e(u)|^{2}dx=0,$

where $c$ depends on $n,$ $\rho_{1}$ and $\epsilon.$ Hence,

$e(u*\rho_{\epsilon})=e(u)*\rho_{\epsilon}\quad\text{a.e. in
$Q^{\prime\prime}.$}$ (A.2)

Recall that $u*\rho_{\epsilon}\in C^{\infty}(Q^{\prime\prime}).$ Since
$e(u)\in L^{2}(Q),$ $e(u)*\rho_{\epsilon}\in C^{\infty}(Q^{\prime\prime})\cap
L^{2}(Q^{\prime\prime})$ in particular, $e(u*\rho_{\epsilon})\in
C^{\infty}(Q^{\prime\prime})\cap L^{2}(Q^{\prime\prime}).$ By Poincaré-Korn
inequality $u*\rho_{\epsilon}\in H^{1}(Q^{\prime\prime}).$ Since
$e(u)*\rho_{\epsilon}\to e(u)$ in $L^{2}(Q^{\prime\prime})$ as $\epsilon\to
0,$ in view of (A.2) there exists $\epsilon_{0}>0$ such that

$\|e(u*\rho_{\epsilon})\|_{L^{2}(Q^{\prime\prime})}\leq\|e(u)\|_{L^{2}(Q^{\prime\prime})}+1\quad\text{for
all $\epsilon\in(0,\epsilon_{0}).$}$

Moreover, by Poincaré-Korn inequality for any $\epsilon\in(0,\epsilon_{0})$
there exists a rigid displacement $a_{\epsilon}$ such that

$\|u*\rho_{\epsilon}-a_{\epsilon}\|_{H^{1}(Q^{\prime\prime})}\leq
C\|e(u*\rho_{\epsilon})\|_{L^{2}(Q^{\prime\prime})}\leq
C(\|e(u)\|_{L^{2}(Q^{\prime\prime})}+1),$

where $C$ is the Poincaré-Korn constant for a cube. Thus, the family
$\\{u*\rho_{\epsilon}\\}_{\epsilon}$ is uniformly bounded in
$H^{1}(Q^{\prime\prime})$. Since $u*\rho_{\epsilon}\to u$ in
$L^{2}(Q^{\prime\prime}),$ there exists a rigid displacement $a$ such that
$a_{\epsilon}\to a$ in $L^{2}(Q^{\prime\prime}).$ Then
$u*\rho_{\epsilon}-a_{\epsilon}$ weakly converges to $u-a$ in
$H^{1}(Q^{\prime\prime}),$ i.e., $u-a\in H^{1}(Q^{\prime\prime}).$ Since $a$
is linear and $\theta^{\prime\prime}$ is arbitrary, $u\in
H_{\mathrm{loc}}^{1}(Q).$ Now covering $U$ with finitely many cubes of
edgelength $2r$ we get $u\in H_{\mathrm{loc}}^{1}(U).$ ∎

## Acknowledgments

Sh. Kholmatov acknowledges support from the Austrian Science Fund (FWF)
project M 2571-N32. P. Piovano acknowledges support from the Austrian Science
Fund (FWF) project P 29681, from the Vienna Science and Technology Fund (WWTF)
together with the City of Vienna and Berndorf Privatstiftung through Project
MA16-005, and from BMBWF through the OeAD-WTZ project HR 08/2020.

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