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

One can take the completion $\mathcal{L}(\mathbb{R}^n)$ of $\mathcal{B}(\mathbb{R}^n)$ with respect to $\lambda$, ie chuck in all sets contained in borel sets of measure 0.

Otherwise one can 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 $A\subseteq\mathbb{R}^n$, $\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\cap E^\complement)$.

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

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  • $\begingroup$ This is a standard theorem of real analysis, the answer is yes. $\endgroup$ – Crostul Feb 28 '15 at 11:22
  • $\begingroup$ Cool. Do you have a reference? $\endgroup$ – user3589318 Feb 28 '15 at 11:26
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    $\begingroup$ Actually no, I'm sorry. $\endgroup$ – Crostul Feb 28 '15 at 11:27
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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.

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  • $\begingroup$ Thanks got it. I should have thought about it a few seconds more before asking.. $\endgroup$ – user3589318 Feb 28 '15 at 11:44
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    $\begingroup$ @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. $\endgroup$ – Michael Cotton Mar 2 '15 at 17:16
  • $\begingroup$ @MichaelCotton I'm trying to understand your answer and i can't really figure out how it helps... $\endgroup$ – Eran Nov 20 '18 at 18:10

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