Is every closed subset of finite measure contained in an open subset of finite measure?

Question

Could someone will verify my statement: For every locally finite Borel measure on metric space and closed set $F$ with finite measure, there exists open set $U$ such that $F \subset U$ and $U$ has finite measure ?

Answer 1

No, not in general.

My metric space is the disjoint union of uncountably many copies of $\mathbb R$. $$X = \bigsqcup_{t \in T} X_t$$ where $T$ is uncountable and $X_t = \mathbb R$ for all $t$. The metric: two points in the same $X_t$ have distance $\min(|x-y|,1)$, two points in different $X_t$ have distance $1$.
My measure is Lebesgue measure $\mu_t$ on each copy $X_t$ of $\mathbb R$. So for a subset $E \subseteq X$ we can write $E = \bigsqcup_{t \in T} E_t$ where $E_t \subseteq \mathbb R$, and its measure is $$\mu(E) = \sum_{t \in T}\mu_t(E_t).$$

This measure is locally finite. Any point in $X$ lies in exactly one set $X_t$ and the open ball of radius $1/2$ centered there has measure $1$.

But your finiteness property fails. Let the closed set be $$F = \bigsqcup_{t \in T} F_t$$ where $F_t = \{0\}$ for all $t$. Then $\mu(F) = 0$. Let $G \supseteq F$ be an open set. I claim $\mu(G) = +\infty$. Indeed, $$G = \bigsqcup_{t \in T} G_t$$ where for all $t$, the set $G_t$ is an open neighborhood of $\{0\}$. So $\mu(G_t) > 0$ for all $t$. And $\mu(G) = \sum \mu_t(G_t)$ is an uncountable sum of positive numbers. So $\mu(G) = +\infty$.