Proof about "metric measure on a compact metric space" I'm reading 
Edgar - Measure, topology, and fractal geometry - 2008; page 157

Proposition 5.4.3. Let $\mathcal{M}$ be a finite metric measure on a compact metric space $S$. Let $E ⊆ S$ be a Borel set. For any $ε > 0$, there exist a compact set $K$ and an open set $U$ with $U ⊇ E ⊇ K$ and $\mathcal{M}(U \ K) < ε$.

In the proof a closed set $F$ is taken and the following sequence is defined
$$
U_n =\{ x ∈ S : dist(x, F )<1/n\}.
$$
Clearly, $U_n\downarrow$ and $\bigcap U_n=F$.
Then the author says
$$
\lim_{n\to\infty}\mathcal{M}(U_n)=\mathcal{M}(F)
$$
I understand that $U_n\supseteq F$, thus $\lim_{n\to\infty}\mathcal{M}(U_n)\ge\mathcal{M}(F)$, but where the converse comes from?
 A: In general, if we have nested measurable sets $\{ U_{n} \}_{n \in \mathbb{N}}$ such that $U_{n} \supset U_{n+1} \supset U_{n+2} \supset \dots$, and there is an $m \in \mathbb{N}$ such that $\mathcal{M}(U_{m}) < \infty$, then it follows that $\lim_{n \rightarrow \infty} \mathcal{M}(U_{n}) = \mathcal{M}(\cap_{n \in \mathbb{N}} U_{n})$. Since the $U_{n}$ as defined satisfies that $\mathcal{M}(U_{1}) \leq \mathcal{M}(S) < \infty$, and $\cap_{n \in \mathbb{N}} U_{n} = F$, it follows that $\lim_{n \rightarrow \infty} \mathcal{M}(U_{n}) = \mathcal{M}(\cap_{n \in \mathbb{N}} U_{n}) = \mathcal{M}(F)$.  
The proof of the first sentence is as follows: Define $G_{n} = U_{n} \setminus U_{n+1}$. Then $U_{1} = (\cap_{n \in \mathbb{N}} U_{n}) \cup (\cup_{n \in \mathbb{N}}G_{n})$. Let $U = \cap_{n \in \mathbb{N}} U_{n}$. Since the $G_{n}$'s are disjoint, we obtain that $\mu(U_{1}) = \mu(U) + \sum_{n \in \mathbb{N}} \mu(U_{n}) - \mu(U_{n+1}) = \mu(U) + \mu(U_{1}) - \lim_{n \rightarrow \infty} \mu(U_{n})$. Since $\mu(U_{1}) < \infty$, we may subtract on both sides to obtain that $\mu(\cap_{n \in \mathbb{N}} U_{n} ) = \lim_{n \rightarrow \infty} \mu(U_{n})$. 
Edit: 
The proof I have given also applies to this example because $U_{n}$ is an open set and Borel Sets are measurable in the Caratheodory sense with respect to the metric outer measure. It will be sufficient to show that $\forall F \subset S$, $F$ closed, $\forall A \subset X$, $\mathcal{M}(A) = \mathcal{M}(A \cap F) + \mathcal{M}(A \setminus F)$, where $\mathcal{M}$ is the metric outer measure. Define $A_{n} = \{ x \in A : d(x,F) \geq \frac{1}{n} \}$. Observe that $A \setminus F = \cup_{n \in \mathbb{N}} A_{n} $. Note that $\mathcal{M}(A) \leq \mathcal{M}(A \cap F) + \mathcal{M}(A \setminus F)$ by monotonicity. Observe that for fixed $n \in \mathbb{N}$, since $A_{n}$ and $F$ are positively separated, $\mathcal{M}(A \cap F) + \mathcal{M}(A_{n}) = \mathcal{M}((A \cap F) \cup A_{n}) \leq \mathcal{M}(A)$. We would like to show that $\mathcal{M}(A \setminus F) \leq \lim_{n \rightarrow \infty} \mathcal{M}(A_{n})$. Assuming this, we then would have that $\mathcal{M}(A \cap F) + \mathcal{M}(A \setminus F) \leq \lim_{n \rightarrow \infty} (\mathcal{M}(A \cap F) + \mathcal{M}(A_{n})) = \lim_{n \rightarrow \infty} \mathcal{M}((A \cap F) \cup A_{n}) \leq \mathcal{M}(A)$, which would imply that $F$ is caratheodory measurable. Define $B_{n} = A_{n} \setminus A_{n-1}$, and $B_{1} = A_{1}$. Then $\mathcal{M}(\cup_{n=k}^{\infty} B_{2n-1}) = \sum_{n=k}^{\infty} \mathcal{M}(B_{2n-1})$, and $\mathcal{M}(\cup_{n=k}^{\infty} B_{2n}) = \sum_{n=k}^{\infty} \mathcal{M}(B_{2n})$, since they are positively separated. Note that $\mathcal{M}(\cup_{n=k}^{\infty} B_{2n}) , \mathcal{M}(\cup_{n=k}^{\infty} B_{2n-1}) \leq \mathcal{M}(A_{2k-1})$ by monotonicity. Assume $\lim_{n \rightarrow \infty} \mathcal{M}(A_{n}) < \infty$, since if it were infinite then $\mathcal{M}(A) = \infty$ and there is nothing to prove. Then $\mathcal{M}(\cup_{n=k}^{\infty} B_{2n}) , \mathcal{M}(\cup_{n=k}^{\infty} B_{2n-1}) < \infty$. Observe that $(A \setminus F) = \cup_{n \in \mathbb{N}} A_{n} \implies \mathcal{M}(A \setminus F) = \mathcal{M}(\cup_{n \in \mathbb{N}} A_{n}) \leq \mathcal{M}(A_{k} \cup (\cup_{n=k+1}^{\infty} B_{n}))$ for any fixed $k \in \mathbb{N}$. Then $\mathcal{M}(A \setminus F) \leq \lim_{k \rightarrow \infty} \mathcal{M}(A_{k})  + \sum_{n=k+1}^{\infty} \mathcal{M}(B_{n})$. Since $\sum_{n=1}^{\infty} \mathcal{M}(B_{n}) < \infty \implies \lim_{k \rightarrow \infty} \sum_{n=k}^{\infty} \mathcal{M}(B_{n}) = 0$. Hence, $\mathcal{M}(A \setminus F) \leq \lim_{n \rightarrow \infty} \mathcal{M}(A_{n}) \implies \mathcal{M}(A \cap F) + \mathcal{M}(A \setminus F)  \leq \mathcal{M}(A \cap F) + \lim_{n \rightarrow \infty} \mathcal{M}(A_{n}) \leq \lim_{n \rightarrow \infty}\mathcal{M}((A \cap F) \cup A_{n}) \leq \mathcal{M}(A)$. Hence, $F$ is caratheodory measurable. This implies that the Borel sets of a metric space are measurable with respect to the metric outer measure. By the Caratheodory extension theorem, $\mathcal{M}$ is countably additive on the borel sets. The proof given earlier now applies to the metric outer measure. 
