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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 $\lambda$.

  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?

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

2 Answers 2

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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.. $\endgroup$
    – Blunka
    Commented Feb 28, 2015 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. $\endgroup$ Commented Mar 2, 2015 at 17:16
  • $\begingroup$ @MichaelCotton I'm trying to understand your answer and i can't really figure out how it helps... $\endgroup$
    – Eran
    Commented Nov 20, 2018 at 18:10
  • $\begingroup$ @Eran my answer develops the details of michael's one, it may be useful to you. $\endgroup$ Commented Jul 31, 2019 at 11:22
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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 $\mathcal{L}=\mathcal{L'}$.

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