TITLE: Prove that if $N$ is a null set in $\mathbb{R}^n$, then there exists a Borel null set $N'$ such that $N\subset N'$.
QUESTION [5 upvotes]: Prove that if $N$ is a null set in $\mathbb{R}^n$, then there exists a Borel null set $N'$ such that $N\subset N'$. In fact, prove that $N'$ may be chosen to be a $G_{\delta}$, a countable intersection of open sets.
So we know $\lambda(N)=0$ by definition of null set ($\lambda$ is Lebesgue measure). I think this theorem might be helpful:
Suppose $A$ is a measurable set in $\mathbb{R}^n$. Then $A$ can be decomposed in the following manner: $A=E\cup N$, $E$ and $N$ are disjoint, $E$ is a Borel set, $N$ is a null set.
Thank you.

REPLY [4 votes]: If $N$ is a null set, then for each integer $k$, there is an open set $G_k$ such that $N \subseteq G_k$ and $|G_k| \le 2^{-k}$.
Define $N' = \bigcap_k G_k$; this is certainly a $G_{\delta}$ set, it contains $N$, and it has measure at most $2^{-k}$ for each integer $k$, and so must have measure $0$.