# 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 $$\lambda$$.

1. 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 $$0$$.

2. 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?

• This is a standard theorem of real analysis, the answer is yes. Feb 28, 2015 at 11:22
• Cool. Do you have a reference? Feb 28, 2015 at 11:26
• Actually no, I'm sorry. Feb 28, 2015 at 11:27

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.

• Thanks got it. I should have thought about it a few seconds more before asking.. Feb 28, 2015 at 11:44
• @user3832080 No worries. I think it's actually a good habit to ask around. And the connection between being measurable and almost $G_\delta$ is one of those many useful little tricks that are easy to forget. Mar 2, 2015 at 17:16
• @MichaelCotton I'm trying to understand your answer and i can't really figure out how it helps...
– Eran
Nov 20, 2018 at 18:10
• @Eran my answer develops the details of michael's one, it may be useful to you. Jul 31, 2019 at 11:22

Just filling the details of Michael's answer:

If $$A\in\mathcal{L}'$$, then consider the family $$\mathcal{C}$$ of every sequence of products of open intervals $$\{R_i\}_{i=1}^{\infty}$$ such that $$A\subset\bigcup_{i=1}^{\infty}R_i$$. Now, by definition of the Lebesgue outer measure, $$\lambda^*(A):=\inf\left\{\sum_{i=1}^{\infty}\mathrm{vol}(R_i):\{R_i\}_{i=1}^{\infty}\in\mathcal{C}\right\}$$. So it is easy to show that there is a decreasing sequence of sets $$\{U_j\}_{j=1}^{\infty}$$ being $$U_j=\bigcup_{i=1}^{\infty}R_i$$ for some $$\{R_i\}_{i=1}^{\infty}\in\mathcal{C}$$ such that $$A\subset B:=\lim_{j\to\infty}U_j:=\bigcap_{j=1}^{\infty}U_j\in\mathcal{B}$$ holds that $$\lambda^*(B)=\lambda^*(A)$$.

Using that $$A\in\mathcal{L}'$$ it follows that $$\lambda^*(B)=\lambda^*(A\cap B)+\lambda^*(B\setminus A)$$, so $$\lambda^*(B\setminus A)=0$$.

Knowing this, we have to proof that given any $$A\in \mathcal{L}'$$, then $$A=A_1\cup A_2$$, being $$A_1\in \mathcal{B}$$ and $$A_2\subset N\in\mathcal{B}$$ such that $$\lambda(N)=0$$. It is easy to show that $$\mathbb{R}^n\setminus A=(B\setminus A)\cup(\mathbb{R}^n\setminus B)$$, which fulfill all our requirements. So, as every $$A\in\mathcal{L'}$$ is the complementary of another set in $$\mathcal{L'}$$, we have won, and $$\mathcal{L}=\mathcal{L'}$$.