Proof. For a Borel integrable function $f\colon [0,\ell] \to \mathbb{R} \cup \{+\infty\}$ we define the auxiliary quantity

\begin{equation*} R_t'(f,n) := \sum_{i=0}^{n-2}\frac{f\left(\ell\left(\frac{t+i}{n}\right)\right)+f\left(\ell\left(\frac{t+(i+1)}{n}\right)\right)}{2}\cdot\frac{\ell}{n}, \end{equation*}

which is the middle term in the definition of $R_t(f,n)$. First of all we estimate

(7)\begin{equation} \begin{aligned} \int_0^1\left\vert R_t(f,n) - R_t'(f,n) \right\vert\,{\mathrm d} t &\le \frac{\ell}{n}\!\int_0^1\!\left\vert \frac{f(0) + f\left(\ell\left(\frac{t}{n}\right)\right)}{2} \right\vert\ + \left\vert \frac{f\left(\ell \left(\frac{t+n-1}{n}\right)\right) + f(\ell)}{2} \right\vert\,{\mathrm d} t \\ &\le \frac{\ell}{2n}\left( f(0) + f(\ell)\right) + \frac{1}{2}\\ &\quad\times\left(\int_0^{\frac{\ell}{n}} f(u)\,{\mathrm d} u + \int_{\ell\left(\frac{n-1}{n}\right)}^\ell f(u)\,{\mathrm d} u \right), \end{aligned} \end{equation}

and the last two terms go to 0 as n goes to $\infty$, respectively because f assume finite values at 0 and $\ell$ and dominated convergence. We now set

\begin{equation*} \begin{aligned} D(f,n):=\int_0^1 \left| R_t'(f,n) -\int_0^\ell f(s)\,{\mathrm d} s \right|\,{\mathrm d} t. \end{aligned} \end{equation*}

If $f \in C^0([0,\ell])$ we have

(8)\begin{equation} D(f,n) \le \int_0^1 \sum_{i=0}^{n-2} \int_{\ell\left(\frac{i}{n}\right)}^{\ell\left(\frac{i+1}{n}\right)} \left| \frac{f\left(\ell\left(\frac{t+i}{n}\right)\right)-f(s)}{2}\right|+\left| \frac{f\left(\ell\left(\frac{t+i+1}{n}\right)\right)-f(s)}{2}\right|\,{\mathrm d} s\,{\mathrm d} t \le \ell\varepsilon \end{equation}

for $n \gt n(f,\varepsilon)$, by uniform continuity of f on $[0,\ell]$. Finally, given two Borel functions f and f ^ʹ, we estimate

(9)\begin{align} \begin{aligned} |D(f,n)-D(f',n)| & \le \int_0^1 |R_t'(f,n)-R_t'(f',n)|\,{\mathrm d} t + \int_0^1 \left| \int_0^\ell (f-f')(s)\,{\mathrm d} s \right|\,{\mathrm d} t \\ &\le \int_0^1 \left| \sum_{i=0}^{n-2}\frac{(f-f')\left(\ell\left(\frac{t+i}{n}\right)\right)+(f-f')\left(\ell\left(\frac{t+(i+1)}{n}\right)\right)}{2}\cdot\frac{\ell}{n}\right|\,{\mathrm d} t\\ &\quad +\|f-f'\|_{L^1(0,\ell)}\\ & \le \frac{1}{2}\sum_{i=0}^{n-2}\int_0^1 |f-f'|\left( \ell\left(\frac{t+i}{n} \right)\right)\,{\mathrm d} t + \frac{1}{2}\sum_{i=0}^{n-2}\int_0^1 |f-f'|\\ &\quad\times \left( \ell\left(\frac{t+i}{n} \right)\right)\,{\mathrm d} t + \|f-f'\|_{L^1(0,\ell)}\\ &\le \frac{1}{2}\sum_{i=0}^{n-2}\int_{\ell(\frac{i}{n})}^{\ell(\frac{i+1}{n})} |f-f'|\,{\mathrm d} t + \frac{1}{2}\sum_{i=0}^{n-2}\int_{\ell(\frac{i+1}{n})}^{\ell(\frac{i+2}{n})} |f-f'|\,{\mathrm d} t \\ &\quad + \|f-f'\|_{{\rm L}^1(0,\ell)}\\ &\le 2 \|f-f'\|_{L^1(0,\ell)}.\\ \end{aligned}\nonumber\\ \end{align}

By approximating $f \in L^1([0,\ell])$ in L ¹-norm with a sequence $\{f_j\} \subseteq C^0([0,\ell])$, applying the triangular inequality, (8) and (9) we conclude that $\varlimsup_{n \to \infty} D(f,n)=0$. This, together with (7), proves the claim.

Proof. The proof is identical to the one of [Reference Eriksson-Bique18, lemma 2.19], where he uses the notion of λ-integral along chains with parameter λ = 1. We just need to modify it for the $\lambda = \frac12$-integral as we did for the proof of proposition 2.4 with respect to the original proof in [Reference Doob15].

Proof. We apply proposition 2.4 to the function $h=g \circ \gamma$, which is Borel and integrable by assumption and satisfies $h(0),h(L) \lt \infty$. In particular, there exists $t\in [0,1]$ and a subsequence n_j such that

\begin{equation*} \lim_{j\to +\infty} R_t(h,n_j) = \int_0^L h(s)\,{\mathrm d} s \end{equation*}

as noted in remark 2.5. For every j we compute

\begin{align*} R_t(h,n_j) &= \frac{h(0) + h\left(L\left(\frac{t}{n_j}\right)\right)}{2} \cdot L\left(\frac{t}{n_j}\right) + \sum_{i=0}^{n_j-2}\frac{h\left(L\left(\frac{t+i}{n_j}\right)\right)+h\left(L\left(\frac{t+(i+1)}{n_j}\right)\right)}{2}\\ &\quad \cdot\frac{L}{n_j} + \frac{h\left(L \left(\frac{t+n_j-1}{n_j}\right)\right) + h(L)}{2} \cdot L\left(\frac{1-t}{n_j}\right)\\ &= \frac{g(\gamma(0)) + g\left(\gamma\left(L\left(\frac{t}{n_j}\right)\right)\right)}{2} \cdot \ell\left(\gamma{|_{\left[0,L\left(\frac{t}{n_j}\right)\right]}}\right) \\ &+ \sum_{i=0}^{n_j-2}\frac{g\left(\gamma\left(L\left(\frac{t+i}{n_j}\right)\right)\right)+g\left(\gamma\left(L\left(\frac{t+(i+1)}{n_j}\right)\right)\right)}{2}\cdot\ell\left(\gamma{|_{\left[ L\left(\frac{t+i}{n_j}\right),L\left(\frac{t+(i+1)}{n_j}\right)\right]}}\right) \\ &+ \frac{g\left(\gamma\left(L \left(\frac{t+n_j-1}{n_j}\right)\right)\right) + g(\gamma(L))}{2} \cdot \ell\left(\gamma{|_{\left[L\left(\frac{1-t}{n_j}\right),1\right]}}\right)\\ &\ge \int_{{\sf c}_{t,n_j}} g\\ \end{align*}

where we used in the last inequality that ${\sf d}(\gamma(a),\gamma(b))\le \ell(\gamma{|_{[a,b]}})$ for every $a,b \in [0,L]$.

Proof. Proposition 4.4 implies that ${\rm F}_{\mathrm{curve}}(u) \le {\rm F}_{\mathscr{C}^{},\,{\rm Lip}}(u)$ for every $u\in \mathcal{L}^{p}({\rm X})$ and this concludes the proof.

Proof of theorem 6.4

For simplicity we divide the proof in steps.

Step 1. It is enough to prove that for every bounded function u with bounded support it holds $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u) \le {\rm F}_{\mathrm{curve}}(u)$. Indeed, by [Reference Heinonen, Koskela, Shanmugalingam and Tyson22, proposition 7.1.35], for every $u\in L^p({\rm X})$ such that ${\rm F}_{\mathrm{curve}}(u) \lt \infty$ we can find a sequence of bounded functions u_j with bounded support such that $u_j \to u$ in $L^p({\rm X})$ and $\varliminf_{j\to +\infty}{\rm F}_{\mathrm{curve}}(u_j) \le {\rm F}_{\mathrm{curve}}(u)$. Therefore

\begin{equation*}\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u) \le \varliminf_{j\to +\infty}\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u_j) \le \varliminf_{j\to +\infty}{\rm F}_{\mathrm{curve}}(u_j) \le {\rm F}_{\mathrm{curve}}(u),\end{equation*}

because of the lower semicontinuity of $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}$. Since $\tilde{\rm F}_{{\rm curve}}$ is the biggest lower semicontinuous functional which is smaller than or equal to ${\rm F}_{\mathrm{curve}}$, we infer that $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u) \le \tilde{\rm F}_{{\rm curve}}(u)$ for every $u\in L^p({\rm X})$ such that ${\rm F}_{\mathrm{curve}}(u) \lt \infty$. The thesis for an arbitrary $u\in L^p({\rm X})$ follows directly by the definitions of $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u)$ and $\tilde{\rm F}_{{\rm curve}}(u)$.

In the following we fix $x_0\in {\rm X}$ and we assume that $u\colon {\rm X} \to [-M,M]$ and that there exists $R\ge 3$ such that $u{|_{{\rm X}\setminus B_R(x_0)}} = 0$.

Step 2. We claim that it is enough to prove the following statement. For every Borel upper gradient $g\in {\rm UG}^{}(u)$ and for every η > 0 there exists another Borel upper gradient $g_\eta \in {\rm UG}^{}(u)$ such that

(14)\begin{equation} \Vert g - g_\eta \Vert_{L^p({\rm X})} \lt \eta, \end{equation}

and with the following property. For every $j\in \mathbb{N}$ there exist functions $u_{\eta,j} \colon {\rm X} \to \mathbb{R}$ and $g_{\eta,j}\in {\rm LUG}^{\frac{1}{j}}(u_{\eta,j})$ such that

(15)\begin{equation} \varlimsup_{j\to +\infty} \Vert u_{\eta,j} - u \Vert_{L^p({\rm X})} \le \eta \quad \text{and} \quad \lim_{j\to +\infty} \Vert g_{\eta,j} - g_\eta \Vert_{L^p({\rm X})} = 0, \end{equation}

Indeed, if the claim is true, then we have

\begin{equation*} \varliminf_{j\to +\infty}{\rm F}_{\mathscr{C}^{},\, {\rm Lip}}(u_{\eta,j}) \le \varliminf_{j\to +\infty} \Vert g_{\eta,j}\Vert_{L^p({\rm X})}= \Vert g_{\eta}\Vert_{L^p({\rm X})} \le \Vert g\Vert_{L^p({\rm X})} + \eta, \end{equation*}

for every η > 0, where we used (14) and (15). By a diagonal argument we deduce that $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u)\le \Vert g\Vert_{L^p({\rm X})}$. By the arbitrariness of $g\in {\rm UG}^{}(u)$ we infer that $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(u) \le {\rm F}_{\mathrm{curve}}(u)$, which is the statement we had to prove from Step 1.

In the remaining steps we will prove the claim of Step 2. In order to simplify the proof we notice that it is enough to prove the statement for upper gradients $g\in {\rm UG}^{}(u)$ that are lower semicontinuous and such that $g \equiv 0$ on ${\rm X}\setminus B_{2R}(x_0)$. The first assertion follows by Vitali-Carathéodory Theorem (cp. [Reference Heinonen, Koskela, Shanmugalingam and Tyson22, page 108]), while the second one follows by truncation: for every $g\in {\rm UG}^{}(u)$, the truncated function $g\cdot \chi_{B_{2R}(x_0)}$ is still an upper gradient of u since $u{|_{{\rm X}\setminus B_R(x_0)}} = 0$ and it is smaller than the original one. In the sequel we assume that g has these properties.

Step 3: For every g as above, we define g_η. By Lusin’s Theorem and the fact that $\mathfrak m$ is Radon, we can find compact sets $K_j \subseteq B_{2R}(x_0)$ such that $\mathfrak m(B_{2R}(x_0) \setminus K_j)^{\frac{1}{p}} \le 2^{-j}\eta$ and so that $u{|_{K_j}}$ is continuous. We can suppose $K_j \subseteq K_{j+1}$ for every j. Let $\sigma \in (0, 1)$ be so that $\mathfrak m(B_{2R}(x_0))^{\frac{1}{p}}\sigma \le \frac{\eta}{2}$. Define

\begin{equation*} g_\eta(x) := g(x) + \sigma\chi_{B_{2R}(x_0)} + \sum_{i=1}^\infty \chi_{B_{2R}(x_0)\setminus K_i}(x). \end{equation*}

The function g_η is still lower semicontinuous and belongs to ${\rm UG}^{}(u)$, since it is bigger than g. Moreover, $g_\eta \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$. We show that (14) holds. Indeed

\begin{equation*} \Vert g - g_\eta \Vert_{L^p({\rm X})} \le \sigma \mathfrak m(B_{2R}(x_0))^{\frac{1}{p}} + \sum_{i=1}^\infty \mathfrak m(B_{2R}(x_0)\setminus K_i)^{\frac{1}{p}} \le \eta. \end{equation*}

In the next step we define auxiliary functions $\hat{g}_{\eta,j}$. Later we will slightly modify these functions in order to define the $g_{\eta,j}$’s of Step 2.

Step 4. We proceed to the definition of $\hat{g}_{\eta,j}$. Let g_j be an increasing sequence of bounded Lipschitz functions such that $g_j \nearrow g$ pointwise, whose existence is guaranteed for instance by [Reference Heinonen, Koskela, Shanmugalingam and Tyson22, corollary 4.2.3]. Define $\psi_{j}(x) := \max\{0,\min\{1, j(2R - {\sf d}(x_0, x))\}\}$. Observe that ψ_j is j-Lipschitz, that $\psi_{j} \le \chi_{B_{2R}(x_0)}$, that $\psi_{j} \equiv 1$ on $B_{2R - \frac{1}{j}}(x_0)$ and that $\psi_{j} \to \chi_{B_{2R}(x_0)}$ pointwise as $j\to +\infty$. We define

\begin{equation*} \hat{g}_{\eta,j}(x) = g_j(x) + \sigma\psi_j(x) + \sum_{i=1}^j \min\{j{\sf d}(x,K_i), \psi_j(x)\}. \end{equation*}

By definition, $\hat{g}_{\eta,j}$ is Lipschitz and bounded. Moreover, $\hat{g}_{\eta,j} \le g_\eta$ for every j and $\hat{g}_{\eta,j} \nearrow g_\eta$ pointwise as $j \to +\infty$.

We define auxiliary functions $\hat{u}_{\eta,j}$. In Step 6 will modify them in order to define the functions $u_{\eta,j}$ required by Step 2.

Step 5. We define $\hat{u}_{\eta,j}$. We choose $N \in \mathbb{N}$ so that $\mathfrak m(B_{2R}(x_0) \setminus K_N)^{\frac{1}{p}} \le (2M)^{-1}\eta$. We define the closed set $A := K_N \cup ({\rm X} \setminus B_R(x_0))$. Since $u{|_{K_N}}$ and $u{|_{X\setminus B_R(x_0)}}$ are continuous and since both sets are closed, then $u{|_{A}}$ is continuous as well. We set

\begin{equation*}\hat{u}_{\eta,j}(x) := \min\left\{M, \inf\left\{u(\alpha({\sf c})) + \int_{\sf c} \hat{g}_{\eta,j} \,:\, {\sf c} \in \mathscr{C}^{\frac{1}{j}}, \omega({\sf c}) = x, \alpha({\sf c}) \in A\right\} \right\}.\end{equation*}

These functions satisfy the following properties:

1. (a) $\hat{u}_{\eta,j}\colon {\rm X} \to [-M,M]$ and $\hat{u}_{\eta,j} \le u$ on A: this follows directly from the definition.

2. (b) $\hat{u}_{\eta,j}$ is $\max\{2Mj, \sup_{\rm X} \hat{g}_{\eta,j}\}$-Lipschitz. Indeed if $x,y\in {\rm X}$ are such that ${\sf d}(x,y) \gt \frac{1}{j}$ then $\vert \hat{u}_{\eta,j}(x) - \hat{u}_{\eta,j}(y)\vert \le 2M \le 2Mj{\sf d}(x,y)$. On the other hand, if ${\sf d}(x,y) \lt \frac{1}{j}$ then $\{x,y\}\in\mathscr{C}^{\frac{1}{j}}_{x,y}$, implying that $\vert \hat{u}_{\eta,j}(y) - \hat{u}_{\eta,j}(x)\vert \le \frac{\hat{g}_{\eta,j}(x) + \hat{g}_{\eta,j}(y)}{2}{\sf d}(x,y) \le \sup_{\rm X} \hat{g}_{\eta,j} {\sf d}(x,y)$.

3. (c) $\hat{g}_{\eta,j} \in {\rm UG}^{\frac{1}{j}}(\hat{u}_{\eta,j})$. We prove that $\hat{u}_{\eta,j}(y) - \hat{u}_{\eta,j}(x) \le \int_{{\sf c}}\hat{g}_{\eta,j}$ for every $x,y\in {\rm X}$ and every ${\sf c} \in \mathscr{C}^{\frac{1}{j}}_{x,y}$. This is enough to show the thesis, since the integral is symmetric. If $\hat{u}_{\eta,j}(x) = M$ there is nothing to prove. Otherwise for every δ > 0 we can find a chain ${\sf c}_\delta \in \mathscr{C}^{\frac{1}{j}}$ with $\omega({\sf c}_\delta) = x$ and $\alpha({\sf c}_\delta) \in A$ such that $\hat{u}_{\eta,j}(x) \ge u(\alpha({\sf c}_\delta)) + \int_{{\sf c}_\delta} \hat{g}_{\eta,j} - \delta$. The chain ${\sf c}_{\delta} \star {\sf c}$ is admissible for the computation of the infimum in the definition of $\hat{u}_{\eta,j}(y)$, giving $\hat{u}_{\eta,j}(y) \le u(\alpha({\sf c}_\delta)) + \int_{{\sf c}_\delta} \hat{g}_{\eta,j} + \int_{\sf c} \hat{g}_{\eta,j} \le \hat{u}_{\eta,j}(x) + \int_{\sf c} g_{\eta,j} + \delta$. The thesis follows by the arbitrariness of δ.

4. (d) $\hat{u}_{\eta,j}(x) \le \hat{u}_{\eta,k}(x)$ if $j\le k$. This follows since $\hat{g}_{\eta,j} \le \hat{g}_{\eta,k}$ and since each $\frac{1}{k}$-chain is also a $\frac{1}{j}$-chain.

5. (e) $\hat{u}_{\eta,j}$ is constant on each $\frac{1}{j}$-chain connected component of ${\rm X} \setminus B_{2R}(x_0)$. Indeed, let $x,y \in {\rm X}$ be such that there exists a $\frac{1}{j}$-chain ${\sf c}_{x,y} \subseteq {\rm X} \setminus B_{2R}(x_0)$. Let ${\sf c} \in \mathscr{C}^{\frac{1}{j}}$ be such that $\alpha({\sf c}) \in A$ and $\omega({\sf c}) = x$. Then ${\sf c}' = {\sf c} \star {\sf c}_{x,y} \in \mathscr{C}^{\frac{1}{j}}$ and satisfies $\alpha({\sf c}') \in A$, $\omega({\sf c}') = y$. Moreover,

   \begin{equation*}\int_{{\sf c}'} \hat{g}_{\eta,j} = \int_{{\sf c}} \hat{g}_{\eta,j} + \int_{{\sf c}_{x,y}} \hat{g}_{\eta,j} = \int_{{\sf c}} \hat{g}_{\eta,j}\end{equation*}

   since $\hat{g}_{\eta,j} \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$. This is enough to show that $\hat{u}_{\eta,j}(y) \le \hat{u}_{\eta,j}(x)$. Reversing the roles of x and y we get the opposite inequality and so that $\hat{u}_{\eta,j}(y) = \hat{u}_{\eta,j}(x)$.

Step 6. Definition of $u_{\eta,j}$ and $g_{\eta,j}$. We define $u_{\eta,j}$ with a cutoff procedure to impose that $u_{\eta,j} \equiv 0$ outside $B_{3R}(x_0)$. We piecewisely define $u_{\eta,j}$. On $B_{2R}(x_0)$ we set $u_{\eta,j} = \hat{u}_{\eta,j}$. Then we define $u_{\eta,j}$ on each $\frac{1}{j}$-chain connected component ${\rm Y}$ of ${\rm X}\setminus B_{2R}(x_0)$ in the following way. By items (e) and (a) of Step 5, we have that $\hat{u}_{\eta,j}$ is constantly equal to some $\delta_{{\rm Y}} \in [-M,0]$ on ${\rm Y}$. We define $u_{\eta,j}$ on ${\rm Y}$ by

\begin{equation*} u_{\eta,j}(x) := \begin{cases} -\frac{\delta_{\rm Y}}{R}{\sf d}(x,x_0) +3\delta_{\rm Y} &\text{if } {\sf d}(x,x_0)\in [2R,3R],\\ 0 &\text{if } {\sf d}(x,x_0) \ge 3R. \end{cases} \end{equation*}

The same proof of item (b) of Step 5 implies that $u_{\eta,j}$ is $\max\{2Mj, \sup_{\rm X} \hat{g}_{\eta,j} + \frac{M}{R}\}$-Lipschitz. Indeed, the only non-trivial case is when ${\sf d}(x,y) \lt \frac{1}{j}$. In that case, if $x,y \in B_{2R}(x_0)$ then the proof does not change. If $x,y \in {\rm X} \setminus B_{2R}(x_0)$ then they must be in the same $\frac{1}{j}$-chain connected component ${\rm Y}$ of ${\rm X} \setminus B_{2R}(x_0)$, so $\vert u_{\eta,j}(x) - u_{\eta,j}(y) \vert \le \frac{\vert \delta_Y \vert}{R}\vert {\sf d}(x,x_0) - {\sf d}(y,x_0) \vert \le \frac{M}{R} {\sf d}(x,y)$. In the last case we have $x\in B_{2R}(x_0)$ and $y\in {\rm X}\setminus B_{2R}(x_0)$. Here we have

\begin{align*}\vert u_{\eta,j}(x) - u_{\eta,j}(y) \vert& \le \vert \hat{u}_{\eta,j}(x) - \hat{u}_{\eta,j}(y) \vert + \frac{\vert \delta_{\rm Y} \vert}{R} ({\sf d}(y,x_0) - 2R) \\ &\le \sup_{{\rm X}} \hat{g}_{\eta,j} {\sf d}(x,y) + \frac{M}{R} {\sf d}(x,y),\end{align*}

where ${\rm Y}$ is the $\frac{1}{j}$-chain connected component of ${\rm X} \setminus B_{2R}(x_0)$ containing y.

It remains to define the new gradients $g_{\eta,j}$. We set

\begin{equation*}\delta_j := \sup \left\{\vert \delta_{\rm Y} \vert \,:\, {\rm Y} \in \mathscr{C}^{\frac{1}{j}}\textrm{-cc}({\rm X} \setminus B_{2R}(x_0))\right\}\end{equation*}

and

\begin{equation*}h_{\eta,j} := \frac{\delta_j}{R} \cdot \max\left\{0,\min\left\{1, 5 - \frac{{\sf d}(x_0, x)}{R}\right\}\right\}.\end{equation*}

Observe that $h_{\eta,j} = \frac{\delta_j}{R}$ on $B_{4R}(x_0)$, that $h_{\eta,j}$ is Lipschitz and that $h_{\eta,j} \equiv 0$ on ${\rm X}\setminus B_{5R}(x_0)$. Finally define $g_{\eta,j} := \hat{g}_{\eta,j} + h_{\eta,j}$. We claim that $g_{\eta,j} \in {\rm LUG}^{\frac{1}{j}}(u_{\eta,j})$. By definition, $g_{\eta,j}$ is Lipschitz, so it remains to show it is a $\frac{1}{j}$-upper gradient of $u_{\eta,j}$. Let ${\sf c} \in \mathscr{C}^{\frac{1}{j}}$. We divide ${\sf c}$ in subchains ${\sf c}_i$ such that $\omega({\sf c}_i) = \alpha({\sf c}_{i+1})$ for every i and such that each ${\sf c}_i$ is of one of the following forms:

• • ${\sf c}_i \subseteq B_{2R}(x_0)$;

• • ${\sf c}_i \subseteq A_{2R,3R}(x_0) := \overline{B}_{3R}(x_0) \setminus B_{2R}(x_0)$;

• • ${\sf c}_i \subseteq {\rm X} \setminus B_{3R}(x_0)$

• • ${\sf c}_i = \{x_i,y_i\}$ with $x_i \in B_{2R}(x_0)$ and $y_i \notin B_{2R}(x_0)$ or $x_i \notin B_{2R}(x_0)$ and $y_i \in B_{2R}(x_0)$;

• • ${\sf c}_i = \{x_i,y_i\}$ with $x_i \in B_{3R}(x_0)$ and $y_i \notin B_{3R}(x_0)$ or $x_i \notin B_{3R}(x_0)$ and $y_i \in B_{3R}(x_0)$.

In all these cases we prove that $\vert u_{\eta,j}(\omega({\sf c}_i)) - u_{\eta,j}(\alpha({\sf c}_i)) \vert \le \int_{{\sf c}_i}g_{\eta,j}$. If ${\sf c}_i \subseteq B_{2R}(x_0)$, we use item (c) of Step 5 to get $\vert u_{\eta,j}(\omega({\sf c}_i)) - u_{\eta,j}(\alpha({\sf c}_i)) \vert = \vert \hat{u}_{\eta,j}(\omega({\sf c}_i)) - \hat{u}_{\eta,j}(\alpha({\sf c}_i)) \vert \le \int_{{\sf c}_i} \hat{g}_{\eta,j} \le \int_{{\sf c}_i} g_{\eta,j}$. If ${\sf c}_i \subseteq A_{2R,3R}(x_0)$, then it must be contained in the same $\frac{1}{j}$-chain connected component ${\rm Y}$ of ${\rm X} \setminus B_{2R}(x_0)$. Therefore we have $\vert u_{\eta,j}(\omega({\sf c}_i)) - u_{\eta,j}(\alpha({\sf c}_i)) \vert \le \frac{\vert \delta_Y \vert}{R}\vert {\sf d}(\omega({\sf c}_i),x_0) - {\sf d}(\alpha({\sf c}_i),x_0) \vert \le \frac{\delta_j}{R} {\sf d}(\omega({\sf c}_i),\alpha({\sf c}_i)) \le \int_{{\sf c}_i} h_{\eta,j} \le \int_{{\sf c}_i} g_{\eta,j}$. If ${\sf c}_i \subseteq {\rm X} \setminus B_{3R}(x_0)$ then $ 0 = \vert u_{\eta,j}(\omega({\sf c}_i)) - u_{\eta,j}(\alpha({\sf c}_i)) \vert \le \int_{{\sf c}_i} g_{\eta,j}$. If ${\sf c}_i = \{x_i,y_i\}$ is as in the last two cases, we have

\begin{equation*} \begin{aligned} \vert u_{\eta,j}(y_i) - u_{\eta,j}(x_i) \vert &\le \vert \hat{u}_{\eta,j}(y_i) - \hat{u}_{\eta,j}(x_i)\vert + \frac{\vert \delta_j \vert}{R}\\ &\quad\times \max\{{\sf d}(y_i,\partial A_{2R,3R}(x_0)), {\sf d}(x_i,\partial A_{2R,3R}(x_0))\}\\ &\le \int_{\{x_i,y_i\}} \hat{g}_{\eta,j} + \int_{\{x_i,y_i\}} h_{\eta,j} = \int_{\{x_i,y_i\}} g_{\eta,j}, \end{aligned} \end{equation*}

because $\max\{{\sf d}(x_i,\partial A_{2R,3R}(x_0)), {\sf d}(y_i,\partial A_{2R,3R}(x_0))\} \le {\sf d}(x_i,y_i)$ and item (c) of Step 5. Therefore

\begin{equation*}\vert u_{\eta,j}(\omega({\sf c})) - u_{\eta,j}(\alpha({\sf c})) \vert \le \sum_i \vert u_{\eta,j}(\omega({\sf c}_i)) - u_{\eta,j}(\alpha({\sf c}_i)) \vert \le \sum_i\int_{{\sf c}_i} g_{\eta,j} = \int_{{\sf c}} g_{\eta,j}.\end{equation*}

Step 7. In this step we show that $u_{\eta,j}$ and $g_{\eta,j}$ satisfy (15) if we prove that $u_{\eta,j}$ converges pointwise to u on K_N and $\hat{u}_{\eta,j}$ converges uniformly to $u \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$ as $j\to +\infty$. Indeed, if this is true, we get that δ_j satisfies

(16)\begin{equation} \varlimsup_{j\to +\infty} \delta_j = \varlimsup_{j \to +\infty} \|\hat{u}_{\eta,j}\|_{L^\infty({\rm X}\setminus B_{2 R}(x_0))}= 0. \end{equation}

Therefore we obtain

\begin{equation*} \begin{aligned} &\lim_{j\to +\infty} \Vert u - u_{\eta,j} \Vert_{L^p({\rm X})} \\ &=\lim_{j\to +\infty}\left(\int_{K_N} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m + \int_{B_{2R}(x_0)\setminus K_N} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m\right.\\ & \quad\left. + \int_{{\rm X} \setminus B_{2R}(x_0)} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m \right)^{\frac{1}{p}} \\ &\le 0 + (2M)\mathfrak m(B_{2R}(x_0) \setminus K_N)^{\frac{1}{p}} + 0 \le \eta. \end{aligned} \end{equation*}

We used dominated convergence for the estimate of the first summand and the estimate $\vert u - u_{\eta,j} \vert \le 2M$ since both functions take values on $[-M,M]$ and $\mathfrak m(B_{2R}(x_0) \setminus K_N)^{\frac{1}{p}} \le \eta (2M)^{-1}$ for the second term. For the third term we divided the integral in two parts: on the annulus $A_{2R,3R}(x_0)$ we used again dominated convergence since $\vert u-u_{\eta,j} \vert = \vert u_{\eta,j}\vert \le \delta_j$ on it, and we can use (16), while on ${\rm X} \setminus B_{3R}(x_0)$ we have $\vert u - u_{\eta,j} \vert = 0$. This concludes the first estimate in (15). For the second one we observe that

\begin{equation*} \varlimsup_{j\to +\infty} \Vert g_{\eta,j} - g_\eta \Vert_{L^p({\rm X})} \le \varlimsup_{j\to +\infty} \left( \Vert \hat{g}_{\eta,j} - g_\eta \Vert_{L^p({\rm X})} + \Vert h_{\eta,j} \Vert_{L^p({\rm X})}\right) = 0, \end{equation*}

where we used dominated convergence on the first term, since $\hat{g}_{\eta,j} \to g_\eta$ pointwise almost everywhere and they are supported on $B_{2R}(x_0)$, and the estimate

\begin{equation*}\Vert h_{\eta,j} \Vert_{L^p({\rm X})} \le \frac{\delta_j}{R} \mathfrak m(\overline{B}_{5R}(x_0))^{\frac{1}{p}},\end{equation*}

where the limit superior of the right hand side is 0 because of (16).

In the last two steps we show that $u_{\eta,j}$ converges to u pointwise on K_N and that $\hat{u}_{\eta,j}$ converges uniformly to 0 outside $B_{2R}(x_0)$.

Step 8. We prove that $u_{\eta,j}$ converges pointwise to u on K_N as $j\to +\infty$. We suppose by contradiction that there exists some $x \in K_N$ such that $u_{\eta,j}(x)$ does not converge to u(x) as j goes to $+\infty$. On K_N we have $u_{\eta,j} = \hat{u}_{\eta,j}$, by definition. By item (d) of Step 5, the sequence $\hat{u}_{\eta,j}(x)$ is increasing and so it admits a limit. Moreover, by item (a) of Step 5, $\hat{u}_{\eta,j}(x) \le u(x)$ for every j. So, our assumption means that $\lim_{j\to +\infty} \hat{u}_{\eta,j}(x) \lt u(x)$. Let us fix δ > 0 such that $\lim_{j\to +\infty} \hat{u}_{\eta,j}(x) \le u(x) - \delta$. Since $u(x) \le M$, we get $\hat{u}_{\eta,j}(x) \le M-\delta$ for every j. By definition, we can find chains ${\sf c}_j \in \mathscr{C}^{\frac{1}{j}}$ such that $\omega({\sf c}_j) = x$, $\alpha({\sf c}_j) \in A$ and

(17)\begin{equation} u(\alpha({\sf c}_j)) + \int_{{\sf c}_j} \hat{g}_{\eta,j} \lt u(x) - \frac{\delta}{2} \le M. \end{equation}

We consider two subchains. If ${\sf c}_j = \{q_0^j, \ldots, q_{N_j}^j = x\}$ we define ${\sf c}_j^s := \{q_0^j,\ldots,q_{i_j}^j\}$ and ${\sf c}_j^e := \{q_{k_j}^j,\ldots,q_{N_j}^j\}$, where i_j is the biggest integer i such that the chain $\{q_0^j,\ldots,q_{i}^j \}$ is contained in $B_{2R - \frac{1}{j}}(x_0)$, while k_j is the smallest integer k such that the chain $\{q_{k}^j,\ldots,q_{N_j}^j = x\}$ is contained in $B_{2R - \frac{1}{j}}(x_0)$. Here, the superscript stay for ‘start’ and ‘end’, respectively. Figure 1 represents these subchains in three different exhaustive situations. As $x\in K_N \subseteq B_{2R}(x_0)$, we have that $x \in B_{2R - \frac{1}{j}}(x_0)$ for j big enough. For these indices, ${\sf c}_j^e$ is not empty, as it contains at least x. Moreover, $\omega({\sf c}_j^e) = x$ and $\alpha({\sf c}_j^e) \in A$. The last assertion can be proved as follows: either ${\sf c}_j^e = {\sf c}_j$, so $\alpha({\sf c}_j^e) = \alpha({\sf c}_j) \in A$, or ${\sf d}(\alpha({\sf c}_j^e), x_0) \ge 2R - \frac{2}{j} \ge R$, because of the maximality property of ${\sf c}_j^e$, and so $\alpha({\sf c}_j^e) \in {\rm X} \setminus B_{R}(x_0) \subseteq A$. On the other hand, ${\sf c}_j^s$ can be empty, and it is empty if and only if $\alpha({\sf c}_j) \notin B_{2R - \frac{1}{j}}(x_0)$. If ${\sf c}_j^s$ is not empty then either ${\sf c}_j^s = {\sf c}_j$, so $\omega({\sf c}_j^s) = x \in A$, or $\omega({\sf c}_j^s) \notin B_{2R - \frac{2}{j}}(x_0)$ by the maximality property of ${\sf c}_j^s$, so $\omega({\sf c}_j^s) \in A$. Moreover, $\alpha({\sf c}_j^s) = \alpha({\sf c}_j) \in A$. In any case the four points $\alpha({\sf c}_j^s)$, $\omega({\sf c}_j^s)$, $\alpha({\sf c}_j^e)$, $\omega({\sf c}_j^e)$ belong to A. There are three possible cases:

1. (1) $\alpha({\sf c}_j) \notin B_{R}(x_0)$, so $u(\alpha({\sf c}_j)) = 0 = u(\alpha({\sf c}_j^e))$. In this case, (17) implies that

   (18)\begin{equation} u(\alpha({\sf c}_j^e)) + \int_{{\sf c}_j^e} \hat{g}_{\eta,j} \le u(\alpha({\sf c}_j)) + \int_{{\sf c}_j} \hat{g}_{\eta,j} \lt u(x) - \frac{\delta}{2} = u(\omega({\sf c}_j^e)) - \frac{\delta}{2}. \end{equation}

2. (2) $\alpha({\sf c}_j) \in B_{R}(x_0)$, which means ${\sf c}_j^s \neq \emptyset$ and $\alpha({\sf c}_j^s) = \alpha({\sf c}_j)$, and we have

   (19)\begin{equation} u(\alpha({\sf c}_j^s)) + \int_{{\sf c}_j^s} \hat{g}_{\eta,j} \lt u(\omega({\sf c}_j^s)) - \frac{\delta}{4}. \end{equation}

3. (3) $\alpha({\sf c}_j) \in B_{R}(x_0)$ and (19) does not hold. In this case, in view of (17), we necessarily have ${\sf c}_j^s \neq {\sf c}_j \neq {\sf c}_j^e$, so $\omega({\sf c}_j^s), \alpha({\sf c}_j^e) \notin B_{2R - \frac{2}{j}}(x_0)$. Moreover

   (20)\begin{equation} \begin{aligned} u(\alpha({\sf c}_j^e)) + \int_{{\sf c}_j^e} \hat{g}_{\eta,j} = u(\omega({\sf c}_j^s)) + \int_{{\sf c}_j^e} \hat{g}_{\eta,j} &\le u(\alpha({\sf c}_j^s)) + \int_{{\sf c}_j^s} \hat{g}_{\eta,j} + \int_{{\sf c}_j^e} \hat{g}_{\eta,j} + \frac{\delta}{4}\\ &\le u(\alpha({\sf c}_j)) + \int_{{\sf c}_j} \hat{g}_{\eta,j} + \frac{\delta}{4} \\ & \lt u(x) - \frac{\delta}{4} = u(\omega({\sf c}_j^e)) - \frac{\delta}{4} \end{aligned} \end{equation}

Figure 1.

The picture shows the definition of ${\sf c}_j^s$ and ${\sf c}_j^e$ in three different situations that cover all possible cases. On the left, $\alpha({\sf c}_j) \notin B_{2R-\frac{1}{j}}(x_0)$, so ${\sf c}_j^s = \emptyset$ and ${\sf c}_j^e \neq {\sf c}_j$. In the middle, $\alpha({\sf c}_j) \in B_{2R-\frac{1}{j}}(x_0)$ and ${\sf c}_j$ is contained in $B_{2R-\frac{1}{j}}(x_0)$, so ${\sf c}_j = {\sf c}_j^s = {\sf c}_j^e$. On the right, $\alpha({\sf c}_j) \in B_{2R-\frac{1}{j}}(x_0)$, but ${\sf c}_j \cap ({\rm X} \setminus B_{2R - \frac{1}{j}}(x_0)) \neq \emptyset$, so ${\sf c}_j^s \neq \emptyset$, ${\sf c}_j^s \neq {\sf c}_j$ and ${\sf c}_j^e \neq {\sf c}_j$.

One of the three cases (1), (2) or (3) holds true for infinitely many j’s. We now show how to conclude the proof supposing that case (1) occurs infinitely many times. Later we will show how to conclude in the other two cases. We restrict to the indices where (1) holds true and we do not relabel the subsequence. We claim that the assumptions of proposition 2.2 (see also the discussion in remark 2.3) are satisfied by $\{{\sf c}_j^e \}_j$.

• • Length. $\ell({\sf c}_j^e) \le \sigma^{-1}\int_{{\sf c}_j^e}\hat{g}_{\eta,j} \le \sigma^{-1}M$, where we used (18), (17) and the fact that ${\sf c}_j^e \subseteq B_{2R - \frac{1}{j}}(x_0)$, so $\hat{g}_{\eta,j} \ge \sigma$ on ${\sf c}_j^e$.

• • Diameter. Since $u{|_{A}}$ is continuous, we can take $\Delta \gt 0$ so that for every $y\in A$ such that ${\sf d}(x, y) \le \Delta$ we have $\vert u(x) - u(y) \vert \le \frac{\delta}{4}$. Since by (18) $u(\alpha({\sf c}_j^e)) \lt u(x) - \frac{\delta}{4}$, and since $\alpha({\sf c}_j^e) \in A$, we conclude that ${\rm Diam}({\sf c}_j^e) \ge {\sf d}(\alpha({\sf c}_j^e), x) \ge \Delta$.

• • h-sum. By definition, $h_j{|_{B_{2R - \frac{1}{j}}(x_0)}} \le \hat{g}_{\eta,j}{|_{B_{2R - \frac{1}{j}}(x_0)}}$, so

  \begin{equation*} \int_{{\sf c}_j^e}h_j \le \int_{{\sf c}_j^e} \hat{g}_{\eta,j}\le M, \end{equation*}

  again by (18) and (17).

Therefore the chains ${\sf c}_j^e$ subconverge to a curve γ of ${\rm X}$, by proposition 2.2 and remark 2.3. We relabel the sequence accordingly, and we denote the chains again by ${\sf c}_j^e$. Since A is closed, so $\alpha(\gamma) = \lim_{j\to +\infty} \alpha({\sf c}_j^e)$ belongs to A, and $u{|_{A}}$ is continuous, we have that $u(\alpha(\gamma)) = \lim_{j\to +\infty} u(\alpha({\sf c}_j^e)).$ This, together with Step 4, says that we are in position to apply lemma 2.7 to the functions $\hat{g}_{\eta,j} \nearrow g_\eta$ and to the sequence of chains ${\sf c}_j^e$ converging to γ, concluding that

\begin{equation*} \begin{aligned} u(\alpha(\gamma)) + \int_\gamma g_\eta \le \varliminf_{j\to +\infty} u(\alpha({\sf c}_j^e)) + \int_{{\sf c}_j^e} \hat{g}_{\eta,j} \le u(x) - \frac{\delta}{2} = u(\omega(\gamma)) - \frac{\delta}{2}. \end{aligned} \end{equation*}

Here we used (18) in the last inequality. This contradicts the fact that $g_\eta \in {\rm UG}^{}(u)$.

Suppose now that case (3) occurs for infinitely many indices j. Then (20) and the same proof given above says that $\{{\sf c}_j^e\}_j$ is a sequence of chains satisfying again the assumptions of proposition 2.2. The remaining part of the argument is exactly the same, using (20) to violate the fact that g_η is an upper gradient of u.

Finally suppose that (2) occurs for infinitely many indices. Then the sequence $\{{\sf c}_j^s\}_j$ satisfies the assumptions of proposition 2.2. Indeed, the estimate of the length and the h-sum is identical, using (19) instead of (18). In the proof of the lower bound of the diameter we need to distinguish two cases. If ${\sf c}_j^s = {\sf c}_j$ then the same proof as above says that ${\rm Diam}({\sf c}_j^s) \ge \Delta$. Otherwise we have that $\omega({\sf c}_j^s) \notin B_{2R-\frac{2}{j}}(x_0)$, so ${\rm Diam}({\sf c}_j^s) \ge {\sf d}(\alpha({\sf c}_j^s), \omega({\sf c}_j^s)) \ge R - \frac{2}{j} \ge 1$. In any case, ${\rm Diam}({\sf c}_j^s) \ge \min\{\Delta, 1\} \gt 0.$ The remaining part of the argument is again the same, using (19) to violate the fact that g_η is an upper gradient of u. We remark that the argument works since both endpoints of ${\sf c}_j^s$ belong to A, which is closed and on which u is continuous.

Step 9. We prove that $\hat{u}_{\eta,j}$ converges uniformly to 0 on ${\rm X}\setminus B_{2R}(x_0)$ as $j\to +\infty$. Suppose it is not the case and recall that $\hat{u}_{\eta,j} \le u$ because of item (a) of Step 5. Then we can find $0 \lt \delta \lt M$ and a sequence of points $x_j\notin B_{2R}(x_0)$ such that $\hat{u}_{\eta,j}(x_j) \lt -\delta$. By definition of $\hat{u}_{\eta,j}$ there must be a chain ${\sf c}_j = \{q_0^j,\ldots,q_{N_j}^j\} \in \mathscr{C}^{\frac{1}{j}}$ with $\alpha({\sf c}_j) \in A$ and $\omega({\sf c}_j) = x_j$ such that $u(\alpha({\sf c}_j)) + \int_{{\sf c}_j}\hat{g}_{\eta,j} \le -\frac{\delta}{2}$. Since $u{|_{{\rm X} \setminus B_{R}(x_0)}} = 0$, then $\alpha({\sf c}_j)$ must belong to $B_R(x_0)$. Let ${\sf c}_j^s = \{q_0^j,\ldots,q_{i_j}^j\}$ be the subchain of ${\sf c}_j$ with the property that i_j is the biggest integer i such that $\{q_0^j,\ldots,q_{i}^j\} \subseteq B_{2R - \frac{1}{j}}(x_0)$. By maximality we have that $q_{i_j}^j \notin B_{2R - \frac{2}{j}}(x_0)$. Therefore $u(q_{i_j}^j) = u(\omega({\sf c}_j^s)) = 0$. Moreover, we have

\begin{equation*} u(\alpha({\sf c}_j^s)) + \int_{{\sf c}_j^s}\hat{g}_{\eta,j} \le u(\alpha({\sf c}_j)) + \int_{{\sf c}_j}\hat{g}_{\eta,j} \le -\frac{\delta}{2} = u(\omega({\sf c}_j^s)) - \frac{\delta}{2}. \end{equation*}

We now claim that $\{{\sf c}_j^s\}_j$ satisfies the assumptions of proposition 2.2. The proof of the upper bound on the length is the same given in Step 8, since $\hat{g}_{\eta,j} \ge \sigma$ on ${\sf c}_j^s$, so

\begin{equation*}\ell({\sf c}_j^s) \le \sigma^{-1}\int_{{\sf c}_j^s} \hat{g}_{\eta,j} \le \sigma^{-1}\left(-u(\alpha({\sf c}_j^s)) - \frac{\delta}{2}\right) \le \sigma^{-1}M.\end{equation*}

For the diameter we have: ${\rm Diam}({\sf c}_j^s) \ge {\sf d}(\alpha({\sf c}_j^s), \omega({\sf c}_j^s)) \ge R - \frac{2}{j}$, since $\alpha({\sf c}_j^s) \in B_R(x_0)$ and $\omega({\sf c}_j^s) \notin B_{2R - \frac{2}{j}}(x_0)$. Finally,

\begin{equation*}\int_{{\sf c}_j^s} h_j \le \int_{{\sf c}_j^s} \hat{g}_{\eta,j} \le M.\end{equation*}

Using again proposition 2.2 and remark 2.3, we conclude that the sequence of chains $\{{\sf c}_j^s\}_j$ subconverges to a curve γ of ${\rm X}$. Arguing as in Step 8 we deduce that g_η violates the upper gradient inequality on γ since the endpoints of ${\sf c}_j^s$ belong to A and u is continuous on A. This is a contradiction.

Proof. Let $g \in {\rm UG}^\varepsilon(u)$. We verify (21). Given ${\sf c} =\{q_i\}_{i=0}^N \in \mathscr{C}^{\varepsilon}$, we compute

\begin{align*} |&(u \varphi)(\omega({\sf c}))- (u \varphi)(\alpha({\sf c}))| \le \sum_{i=0}^{N-1} |(u \varphi)(q_{i+1})- (u \varphi)(q_{i})|\\ & \le \sum_{i=0}^{N-1} \bigg| u(q_{i+1})\varphi(q_{i+1}) -\frac{1}{2} u(q_{i+1})\varphi(q_{i}) + \frac{1}{2} u(q_{i+1})\varphi(q_{i}) \\ &\qquad\quad-\frac{1}{2}u(q_{i}) \varphi(q_{i+1})+ \frac{1}{2}u(q_{i}) \varphi(q_{i+1}) - u(q_{i}) \varphi(q_{i}) \bigg| \\ &\le \sum_{i=0}^{N-1} \left( \frac{1}{2}\vert u(q_{i+1}) \vert{\rm sl}_\varepsilon \varphi(q_{i+1}) + \frac{1}{2}\vert u(q_{i}) \vert{\rm sl}_\varepsilon \varphi(q_{i}) \right)\,{\sf d}(q_i,q_{i+1})\\ &\qquad\quad+ \frac{1}{2}\vert \varphi(q_{i+1})\vert \vert u(q_{i+1})-u(q_{i}) \vert + \frac{1}{2}\vert \varphi(q_{i}) \vert \vert u(q_{i+1})-u(q_{i}) \vert \\ & \le \int_{{\sf c}} \vert u \vert\,{\rm sl}_{\varepsilon} \varphi + \sum_{i=0}^{N-1} \frac{1}{4} (\vert \varphi(q_{i})\vert + \vert\varphi(q_{i+1})\vert)(g(q_i)+g(q_{i+1})){\sf d}(q_i,q_{i+1})\\ & \le \int_{{\sf c}} \vert u \vert\,{\rm sl}_{\varepsilon} \varphi + \sum_{i=0}^{N-1} \frac{(Q_\varepsilon \varphi\, g)(q_i) + (Q_\varepsilon \varphi\, g)(q_{i+1})}{2} {\sf d}(q_i,q_{i+1}) = \int_{{\sf c}} (\vert u \vert \,{\rm sl}_{\varepsilon} \varphi + Q_\varepsilon \varphi\, g). \end{align*}

Now, to estimate ${\rm F}_{\mathscr{C}^{}}(\varphi\, u)$, we use that $\inf \{\|h\|_{L^p({\rm X})}:\, h \in {\rm UG}^\varepsilon(u \varphi),\, h \text{Borel}\}$ is less than or equal to

\begin{equation*} \begin{aligned} &\inf\{\| |u|\,{\rm sl}_{\varepsilon} \varphi + Q_\varepsilon \varphi\, g \|_{L^p({\rm X})}\,:\, g\in {\rm UG}^{\varepsilon}(u),\, g\ \text{Borel} \} \\ \le &\inf\{\| |u| \,{\rm sl}_{\varepsilon} \varphi \|_{L^p({\rm X})} + \| Q_\varepsilon \varphi\, g \|_{L^p({\rm X})}\,:\, g\in {\rm UG}^{\varepsilon}(u),\, g\ \text{Borel} \}\\ \le &\left(\int |u|^p\,{\rm sl}_{\varepsilon}\varphi^p\,{\mathrm d} \mathfrak m\right)^{\frac{1}{p}} + \| \varphi \|_{L^\infty({\rm X})} \inf\{\|g\|_{L^p({\rm X})}:\, g \in {\rm UG}^\varepsilon(u),\, g\ \text{Borel} \}\\ \end{aligned} \end{equation*}

where we used that $\|Q_\varepsilon \varphi\|_{L^\infty({\rm X})} =\|\varphi\|_{L^\infty({\rm X})}$. By taking the limit as ɛ → 0 and using the fact that φ is Lipschitz, dominated convergence and the definition of ${\rm F}_{\mathscr{C}^{}}(\cdot)$, we get the conclusion.

Proof of theorem 6.7

The proof is a modification of the one of theorem 6.4. We highlight what are the differences in each step.

Step 1. The proof does not change if we show that for every $u\in L^p({\rm X})$ we can find a sequence of bounded functions u_j with bounded support such that $u_j \to u$ in $L^p({\rm X})$ and $\varliminf_{j\to +\infty}{\rm F}_{\mathscr{C}^{}}(u_j) \le {\rm F}_{\mathscr{C}^{}}(u)$. We fix a basepoint $x_0 \in {\rm X}$ and we consider the 1-Lipschitz function $\varphi_j(x) = \max\{0,\min\{1,j+1-{\sf d}(x,x_0)\}\}$. We define $u_j := \min \{j, \max\{-j, \varphi_j u\}\}$. By definition, u_j is bounded and has bounded support. Moreover, every ɛ-upper gradient of $\varphi_j u$ is also a ɛ-upper gradient of u_j. This implies that ${\rm F}_{\mathscr{C}^{}}(u_j) \le {\rm F}_{\mathscr{C}^{}}(\varphi_j u)$ for every j. proposition 6.8 implies that

\begin{equation*} \begin{aligned} \varliminf_{j\to +\infty} {\rm F}_{\mathscr{C}^{}}(u_j) \le \varliminf_{j\to +\infty} {\rm F}_{\mathscr{C}^{}}(\varphi_j u) &\le \varliminf_{j\to +\infty}\left( \int |u|^p ({\rm lip}\,\varphi_j)^p \,{\mathrm d} \mathfrak m \right)^{\frac{1}{p}} + \|\varphi_j\|_{L^\infty}\, {\rm F}_{\mathscr{C}^{}}(u)\\ &= \varliminf_{j\to +\infty}\left( \int_{\overline{B}_{j+1}(x_0)\setminus B_j(x_0)} |u|^p \,{\mathrm d} \mathfrak m \right)^{\frac{1}{p}} +{\rm F}_{\mathscr{C}^{}}(u)\\ & = {\rm F}_{\mathscr{C}^{}}(u), \end{aligned} \end{equation*}

where in the last equality we used that $u\in L^p({\rm X})$.

Step 2. It does not change. In particular the claim we have to prove is the following. For every ɛ > 0, for every Borel ɛ-upper gradient $g\in {\rm UG}^{\varepsilon}(u)$ and for every η > 0 there exists another Borel ɛ-upper gradient $g_\eta \in {\rm UG}^{\varepsilon}(u)$ such that

\begin{equation*} \Vert g - g_\eta \Vert_{L^p({\rm X})} \lt \eta \end{equation*}

and with the following property. For every $j\in \mathbb{N}$ there exist functions $u_{\eta,j} \colon {\rm X} \to \mathbb{R}$ and $g_{\eta,j}\in {\rm LUG}^{\frac{1}{j}}(u_{\eta,j})$ such that

\begin{equation*} \varlimsup_{j\to +\infty} \Vert u_{\eta,j} - u \Vert_{L^p({\rm X})} \le 2\eta \quad \text{and} \quad \lim_{j\to +\infty} \Vert g_{\eta,j} - g_\eta \Vert_{L^p({\rm X})} = 0. \end{equation*}

Let $R\ge 1$ be such that $u\equiv 0$ on ${\rm X}\setminus B_{R}(x_0)$. We recall that it is enough to consider chain upper gradients $g\in {\rm UG}^{\varepsilon}(u)$ that are lower semicontinuous and such that $g \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$, by a truncation argument.

Step 3. The definition of g_η does not change. Observe that $g_\eta \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$ and that $g_\eta \in {\rm UG}^{\varepsilon}(u)$ because $g_\eta \ge g$.

Step 4. The definition of $\hat{g}_{\eta,j}$ does not change and satisfies the same properties.

Step 5. Here there is a difference in the definition of the set A. Since $g_\eta \in L^p({\rm X})$ then $\mathfrak m(\{g_\eta = +\infty \})=0$. By outer regularity of the measure we can find an open set U_η containing $\{g_\eta = +\infty \}$ and such that $\mathfrak m(U_\eta)^{\frac{1}{p}} \lt \eta (2M)^{-1}$. Moreover, since $g_\eta \equiv 0$ on ${\rm X} \setminus B_{2R}(x_0)$, we can choose U_η such that $U_\eta \subseteq B_{2R}(x_0)$. Now we change the definition of the set A by setting $A := (K_N \cup ({\rm X} \setminus (B_R(x_0))) \setminus U_\eta$. It is still closed and $u{|_{A}}$ is still continuous. Now, the definition of $\hat{u}_{\eta,j}$ does not change, except for the fact that we use this set A. Properties (a)-(e) continue to hold.

Step 6. The definitions of $u_{\eta,j}$ and $g_{\eta,j}$ do not change.

Step 7. Here we claim that it is enough to show that $u_{\eta,j}(x)$ converges to u(x) for every $x\in K_N\setminus U_\eta$ as $j \to +\infty$ and that $\hat{u}_{\eta,j}$ converges uniformly to 0 on ${\rm X}\setminus B_{2R}(x_0)$. Indeed if this is true we have

\begin{equation*} \begin{aligned} &\lim_{j\to +\infty} \int_{\rm X} \Vert u - u_{\eta,j} \Vert_{L^p({\rm X})} \,{\mathrm d}\mathfrak m \\ &\le\lim_{j\to +\infty}\left(\int_{K_N \setminus U_\eta} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m + \int_{(B_{2R}(x_0)\setminus K_N) \cup U_\eta} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m\right.\\ &\quad \left.+ \int_{({\rm X} \setminus B_{2R}(x_0))} \vert u - u_{\eta,j} \vert^p \,{\mathrm d}\mathfrak m \right)^{\frac{1}{p}}\\ &\le 0 + (2M) \mathfrak m(B_{2R}(x_0) \setminus K_N)^{\frac{1}{p}} + (2M)\mathfrak m(U_\eta)^{\frac{1}{p}} + 0 \le 2\eta. \end{aligned} \end{equation*}

Step 8. Here we need an additional argument that justifies the different choice of A. Indeed, we claim that in any of three cases, the limit curve has its extreme points in A.

In cases (1) and (3) this is true because either $\alpha({\sf c}_j^e) = \alpha({\sf c}_j) \in A$ by definition, or ${\sf d}(\alpha({\sf c}_j^e), {\rm X} \setminus B_{2R}(x_0)) \le \frac{2}{j}$, by maximality of ${\sf c}_j^e$. In the first case the result is trivial because A is closed and $u{|_{A}}$ is continuous. In the second case $\alpha(\gamma) = \lim_{j\to +\infty} \alpha({\sf c}_j^e) \in {\rm X} \setminus B_{2R}(x_0)$, so $\alpha(\gamma)\in A$ since $U_\eta \subseteq B_{2R}(x_0)$. Moreover, $u(\alpha(\gamma)) = 0 = u(\alpha({\sf c}_j^e))$ because all these points belong to ${\rm X} \setminus B_R(x_0)$. On the other hand $\omega(\gamma) = \lim_{j\to +\infty} \omega({\sf c}_j^e) = x \in A$ because x is chosen in $K_N\setminus U_\eta$. Here, we have $u(\omega(\gamma)) = u(x) = u(\omega({\sf c}_j^e))$.

In case (2) we have that $\alpha(\gamma) = \lim_{j\to +\infty} \alpha({\sf c}_j^s) \in A$, because $\alpha({\sf c}_j^s) \in A$ for every j and A is closed. Moreover, $u(\alpha(\gamma)) = \lim_{j\to +\infty} u(\alpha({\sf c}_j^s))$ since $u{|_{A}}$ is continuous. On the other hand, either $\omega({\sf c}_j^s) = x$ for every j, so $\omega(\gamma) = x \in A$ and $u(\omega(\gamma)) = u(x) = u(\omega({\sf c}_j^s))$, or ${\sf d}(\omega({\sf c}_j^s), {\rm X} \setminus B_{2R}(x_0)) \le \frac{2}{j}$, by maximality of ${\sf c}_j^s$. Arguing as before, we get that $\omega(\gamma) \in {\rm X} \setminus B_{2R}(x_0)$, so it belongs to A and $u(\omega(\gamma)) = 0 = u(\omega({\sf c}_j^s))$.

In every case, the extreme points of γ belong to the set $\{g_\eta \lt +\infty\}$. Hence g_η satisfies the upper gradient inequality along γ because of proposition 4.4, while the proof shows that this is not the case, giving a contradiction.

Step 9. The proof does not change, using the same modifications we did in Step 8.

Proof. Let $u_0\in \mathcal{L}^p(\bar{{\rm X}})$ be any representative of $\iota(u)$. Since ɛ-chains in ${\rm X}$ are ɛ-chains in $\bar{{\rm X}}$, restrictions of $\bar{\mathfrak m}$-measurable elements in ${\rm UG}^{\varepsilon}(u_0)$ are $\mathfrak m$-measurable and belong to ${\rm UG}^{\varepsilon}(u)$ and the property of being Lipschitz is preserved. Thus, $H^{1,p}_{{\rm \mathscr{C}^{}}}(\bar{{\rm X}})\subseteq \iota(H^{1,p}_{{\rm \mathscr{C}^{}}}({\rm X}))$ and $\|u\|_{H^{1,p}_{{\rm \mathscr{C}^{}}}({\rm X})} \le \| \iota(u) \|_{H^{1,p}_{{\rm \mathscr{C}^{}}}(\bar{{\rm X}})}$ for every $u \in L^p({\rm X})$, and similarly $H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\bar{{\rm X}})\subseteq \iota(H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}({\rm X}))$ and $\|u\|_{H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}({\rm X})} \le \| \iota(u) \|_{H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\bar{{\rm X}})}$ for every $u \in L^p({\rm X})$.

For the other inequality we proceed in two different ways. If $u \in L^p({\rm X}) \cap {\rm Lip}({\rm X})$ and $g\in {\rm LUG}^{\varepsilon}(u) \cap L^p({\rm X})$, then we consider the Lipschitz extensions $\bar{u}$, $\bar{g}$ of u and g on $\bar{{\rm X}}$. We claim that $\bar{g}\in {\rm LUG}^{\frac{\varepsilon}{2}}(\bar{u}) \cap L^p({\rm X})$. Indeed, given a chain ${\sf c} = \{q_i\}_{i=0}^N \in \mathscr{C}^{\frac{\varepsilon}{2}}(\bar{{\rm X}})$ we can find sequence of chains ${\sf c}_j = \{q_i^j\}_{i=0}^N \in \mathscr{C}^{\varepsilon}({\rm X})$ such that $q_i^j$ converges to q_i for every $i=0,\ldots,N$ as $j\to +\infty$. By continuity of $\bar{g}$ and $\bar{u}$ we then have

\begin{equation*}\bar{u}(\omega({\sf c})) - \bar{u}(\alpha({\sf c})) = \lim_{j\to +\infty} u(\omega({\sf c}_j)) - u(\alpha({\sf c}_j)) \le \lim_{j\to +\infty}\int_{{\sf c}_j} g = \int_{{\sf c}}\bar{g}.\end{equation*}

Therefore $\tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\iota(u)) = \tilde{\rm F}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\bar{u}) \le {\rm F}_{\mathscr{C}^{},\,{\rm Lip}}(u)$ for every $u\in L^p({\rm X})$. This is enough to conclude that $\iota(H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}({\rm X})) \subseteq H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\bar{{\rm X}})$ and $\|\iota(u) \|_{H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}(\bar{{\rm X}})} \le \|u\|_{H^{1,p}_{{\rm \mathscr{C}^{},\,{\rm Lip}}}({\rm X})}$ for every $u \in L^p({\rm X})$.

For the remaining inequality we recall that in the definition of ${\rm F}_{\mathscr{C}^{}}(u)$ the infimum of the $L^p({\rm X})$-norms can be taken among the p-weak ɛ-upper gradients of u that are $\mathfrak m$-measurable. Moreover, since $\mathfrak m(\bar{{\rm X}}\setminus {\rm X}) = 0$, then ${\mathscr C}\text{-}{\rm Mod}_{p}^{\varepsilon}(\mathscr{C}^{}(\bar{{\rm X}}\setminus {\rm X})) = 0$. This means that every $g\in {\rm WUG}_{p}^{\varepsilon}(u) \cap L^p({\rm X})$ defines a $\bar{\mathfrak m}$-measurable function on $\bar{{\rm X}}$, by (2), which belongs to ${\rm WUG}_{p}^{\varepsilon}(u_0) \cap L^p(\bar{{\rm X}})$, where $u_0\in \mathcal{L}^p(\bar{{\rm X}})$ is any representative of $\iota(u)$. Hence ${\rm F}_{\mathscr{C}^{}}(u_0) \le {\rm F}_{\mathscr{C}^{}}(u)$ for every $u\in L^p({\rm X})$. This implies that $\iota(H^{1,p}_{{\rm \mathscr{C}^{}}}({\rm X})) \subseteq H^{1,p}_{{\rm \mathscr{C}^{}}}(\bar{{\rm X}})$ and $\|\iota(u) \|_{H^{1,p}_{{\rm \mathscr{C}^{}}}(\bar{{\rm X}})} \le \|u\|_{H^{1,p}_{{\rm \mathscr{C}^{}}}({\rm X})}$ for every $u \in L^p({\rm X})$.