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 ? 



Is every closed subset of finite measure contained in an open subset of finite measure?
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 $\mu(G) = +\infty$.


