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A metric measure space is a tuple $(X,d,\mu)$ where $(X,d)$ is a metric space and $\mu$ is a measure on $(X,\mathcal{B}_X)$ ($\mathcal{B}_X$ is the notation for the Borel $\sigma$-algebra associated to the metric space $(X,d)$).

I'm studying functions of bounded variation on metric spaces. The article that I'm using is this one: https://arxiv.org/abs/1810.05310.

Let $(X,d)$ be a metric space and $U\subseteq X$ open. Given a function $f:U\rightarrow\mathbb{R}$, we define its variation on $U$ as follows:

$$V_U(f)=\inf\left\{\liminf_{n\to\infty}{\int_{U}{\text{lip}{f_n}\,d\mu}}\middle|\{f_n\}_{n\in\mathbb{N}}\subseteq\text{Lip}_{\text{loc}}(U),\lim_{n\to\infty}{f_n}=f\text{ on }\mathcal{L}^1_{\text{loc}}(U,\mu)\right\}.$$

They assume that $\text{Lip}_{\text{loc}}(U)$ is dense on $\mathcal{L}^1_{\text{loc}}(U,\mu)$ to this definition. I would like to know a result that, with the minimum hypothesis on the measure metric space, we garantee that's true. If someone coul send a link to an article where they prove or at least mention this result, I'd really appreciated.

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I think the proper thing to check is whether the closure of $\text{Lip}_{\text{loc}}(U)$, in the $\mathcal{L}^1_{\text{loc}}(U,\mu)$ topology, contains all simple functions. Since simple functions determine the integral of any locally integrable function, this is enough.

If you assume your measure is Radon and the space is proper, then the span of indicator functions $\mathbf{1}_F$, where $F$ is a closed subset, is dense in $\mathcal{L}^1_{\text{loc}}(U,\mu)$. You can then approximate $\mathbf{1}_F$ by a sequence of Lipschitz functions, by defining

$$ \phi_n(x):= \begin{cases} 1 & ;x\in F\\ 0 &; d(x,F)\geq \frac{1}{n} \\ 1-n \cdot d(x,F) & ;else \end{cases}. $$

I can not speak to the optimality of these conditions, but given the article you linked to, I think this is sufficient for those cases. I think if your measure is Radon and the span of indicator functions of compact sets is dense in $\mathcal{L}^1_{\text{loc}}(U,\mu)$, this should also probably be enough.

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  • $\begingroup$ @RaúlFiligranaVillalba Do you mean showing $\vert \phi_n(x)-\phi_n(y)\vert \leq K \cdot d(x,y)$ when $0<d(x,F)<\frac{1}{n}$ and $d(y,F)\geq \frac{1}{n}$? $\endgroup$ Apr 12, 2023 at 15:48
  • $\begingroup$ Yes, but I got it. I didn't take into account that if $\phi_n(y)=0$, then $n\cdot d(y,F)\geq 1$ and it follows easily. Thanks. $\endgroup$ Apr 12, 2023 at 20:21

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