This is an exercise question in my Lebesgue Integration course. I feel like I should make use of collection of closed sets but I have no idea where to start.
Here is the question:
Let $E\in M$ be such that $\mu(E)$ = $0$. Show that there is a $B$ in the Borel $\sigma$-algebra of $\mathbb{R}$ such that $E\subset B$ and $\mu(B) = 0$.
Here $(\mathbb{R}, M, \mu)$ is Lebesgue measurable space.
Any help would be appreciated.
Since $E$ is measurable, we know that $0 = \mu(E) = \mu^* (E) = \inf m^*(U)$ where $U$ is an open set containing $E$. Thus, for any $\epsilon > 0$ we can find an open $U \supset E$ such that $\mu(U) < \epsilon$. Thus, for any $n \in \mathbb{N}$, there is an open $O_n \subset \mathbb{R}$ such that $E \subset O_n$ and $\mu (O_n) < \frac{1}{n}$.
Now take the following set: $$O = \bigcap_{n=1}^\infty O_n$$
$O$ is a $G_\delta$ set, and hence is a Borel set. Additionally, $O$ contains $E$ and has measure $0$.