Completion of borel sigma algebra with respect to Lebesgue measure There are two ways of extending the Borel $\sigma$-algebra on $\mathbb{R}^n$, $\mathcal{B}(\mathbb{R}^n)$, with respect to Lebesgue measure $\lambda$.


*

*The completion $\mathcal{L}(\mathbb{R}^n)$ of $\mathcal{B}(\mathbb{R}^n)$ with respect to $\lambda$, i.e. chuck in all sets contained in Borel sets of measure $0$.


*let $\lambda^*$ be outer Lebesgue measure on $\mathcal{P}(\mathbb{R}^n)$, and take $\mathcal{L}'(\mathbb{R}^n)$ to be those $E$ such that for all $A\subseteq\mathbb{R}^n$, $\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^\complement)$.

We know that $\mathcal{L}'(\mathbb{R}^n)\supset\mathcal{B}(\mathbb{R}^n)$ and $ \mathcal L'(\mathbb R^n) $ is complete, so $\mathcal L'(\mathbb R^n)\supset\mathcal{L}(\mathbb{R}^n)$. But does the reverse inclusion also hold?

From the definition of the outer measure $\lambda^{*}$, you can show that if $A\in \mathcal{L}'$ then there's a $G_{\delta}$ set $B$ so that $A\subseteq B$ and $\lambda^{*}(B\setminus A)=0$. After that, the answer to this question is an easy yes.
