Let $(X,d,\mu)$ be a $\sigma$-compact metric measure space such that $\mu$ is finite on compact sets and $U\subseteq X$ an open set. With these hypothesis we know that $\mathcal{L}^1_{\text{loc}}(U,\mu)$ is metrizable with a metric compatible with the convergence on $\mathcal{L}^1_{\text{loc}}(U,\mu)$ and $\text{Lip}_{\text{loc}}(U)$ (also $\text{Lip}(U)$) is dense on that space.
If $f\in\mathcal{L}^1_{\text{loc}}(U,\mu)$ we define the variation of $f$ on $U$ as follows:
$$V_U(f)=\inf\left\{\liminf_{n\to\infty}{\int_{U}{\text{lip }f_n\,d\mu}}\middle|\{f_n\}_{n\in\mathbb{N}}\subseteq\text{Lip}_{\text{loc}}(U),\lim_{n\to\infty}{f_n}=f\text{ on }\mathcal{L}^1_{\text{loc}}(U,\mu)\right\}.$$
It's very easy to prove that on those conditions, there exists a sequence $\{f_n\}_{n\in\mathbb{N}}\subseteq\text{Lip}_{\text{loc}}(U)$ such that $\lim_{n\to\infty}{f_n}=f$ on $\mathcal{L}^1_{\text{loc}}(U,\mu)$ and:
$$V_U(f)=\lim_{n\to\infty}{\int_{U}{\text{lip }{f_n}\,d\mu}}.$$
We define $||Df||:\mathcal{P}(U)\rightarrow[0,+\infty]$ as follows:
$$||Df||(A)=\inf\left\{||Df||(V)\middle|V\subseteq U\text{ open},A\subseteq V\right\}.$$
I want to prove that $||Df||$ defines an outer measure, but I had problems with subadditivity.
Let $\{A_n\}_{n\in\mathbb{N}}$ be a sequence of subsets of $U$ and $A:=\bigcup_{n=1}^{\infty}{A_n}$. If $\sum_{n=1}^{\infty}{||Df||(A_n)}=+\infty$, then clearly $||Df||(A)\leq\sum_{n=1}^{\infty}{||Df||(A_n)}$. Suppose that $\sum_{n=1}^{\infty}{||Df||(A_n)}<+\infty$ and, consequently, $||Df||(A_n)<+\infty$ for all $n\in\mathbb{N}$. Then, fixed $\varepsilon>0$, for each $n\in\mathbb{N}$, there exists an open set $V_n\subseteq U$ such that $A_n\subseteq V_n$ and:
$$||Df||(V_n)<||Df||(A_n)+\dfrac{\varepsilon}{2^n}.$$
Let $V:=\bigcup_{n=1}^{\infty}{V_n}\subseteq U$. Observe that $A\subseteq V$. If we prove subadditivity on open sets, we had the result:
$$||Df||(A)\leq||Df||(V)\leq\sum_{n=1}^{\infty}{||Df||(V_n)}\leq\sum_{n=1}^{\infty}{||Df||(A_n)}+\varepsilon.$$
How could I do it? I tried to use that $||Df||(V_n)$ is limit of integrals as I said above, but I didn't get anything.