Let $X$ be a separable metric space, with $(X, \tau)$ the corresponding topological space, and let $\mu$ be a (positive) measure on $(X, \mathcal{B}(X))$, where $\mathcal{B}(X)$ is the Borel $\sigma-$algebra generated by the metric topology. I am trying to prove the following result.
Let $\mathcal{U} := \{ U \mathrm{~is~open}: \mu(U) = 0 \}.$ Then $\mu \left (\bigcup \limits_{U \in \mathcal{U}} U \right) = 0.$
Attempt:
Because $X$ is second-countable, there exist countably many open sets $U_1, U_2, \ldots$ such that $\bigcup \limits_{U \in \mathcal{U}} U = \bigcup \limits_{n = 1}^\infty U_n.$ It follows that $\mu \left (\bigcup \limits_{U \in \mathcal{U}} U \right ) \leq \sum_{n = 1}^\infty \mu(U_n).$
My question is the following: How do we know that $\mu(U_n) = 0$ for all $n$?
This fact is used without explanation in the textbook "Probability measures on metric spaces" by K.R. Parthasarathy, Chapter 2, in the proof of Theorem 2.1. You can assume that $\mu$ is a probability measure, although I don't see why this would help.
After further research, I found the following:
A separable metric space is a Lindelöf space, that is, a topological space in which every open cover has a countable subcover.
Trivially, $\mathcal{U}$ is an open cover of $\bigcup \limits_{U \in \mathcal{U}} U$. Because $X$ is Lindelöf, it has a countable subcover $\{U_n\}_{n = 1}^{\infty}$ such that $U_n$ is open and $\mu(U_n) = 0$ for all $n$. It follows that $\mu \left (\bigcup \limits_{U \in \mathcal{U}} U \right ) \leq \sum_{n = 1}^\infty \mu(U_n) = 0.$