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This is an exercise in real analysis:

Let $E\subset {\Bbb R}^d$. For every $\varepsilon>0$, one can find a Lebesgue measurable set $E_{\varepsilon}$ such that $m^*(E_{\varepsilon}\Delta E)\leq\varepsilon$. Show that $E$ is Lebesgue measurable.

The definition of Lebesgue measurable used here is

Let $E\subset{\Bbb R}^d$, $E$ is Lebesgue measurable if for any $\varepsilon>0$, there exists an open set $U\supset E$ such that $m^*(U\setminus E)\leq\varepsilon$, where $m^*$ is Lebesgue outer measure.

Directly using the definition above might be difficult. I'm trying to use the fact that the Lebesgue measurable sets form a $\sigma$-algebra. But I don't see the way to write $E$ as a union of Lebesgue measurable set. Any help?

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What is $E_{\epsilon}\triangle E$ ? Is it the symmetric difference ? –  Amr Jan 3 '13 at 16:17
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Yes, it is the symmetric difference. –  Jack Jan 3 '13 at 16:36
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2 Answers

up vote 1 down vote accepted

For any $\varepsilon>0$, there exists a family of L-measurable sets $\{E_{\varepsilon,n}\}_{n=1}^{\infty}$ such that $$ m^*(E_{\varepsilon,n}\Delta E)\leq \varepsilon/2^n $$ Let $$ E_{\varepsilon}=\cup_{n=1}^{\infty}E_{\varepsilon,n}. $$ Then $E_{\varepsilon}$ is L-measurable and $$ E\setminus E_{\varepsilon}\subset E\setminus E_{\varepsilon,n} $$ for all $n$. By monotonicity of Lebesgue outer measure, $$ m^*(E\setminus E_{\varepsilon})\leq m^*( E\setminus E_{\varepsilon,n})\leq \varepsilon/2^n $$ Thus $m^*(E\setminus E_{\varepsilon})=0$ which implies that $E\setminus E_{\varepsilon}$ is L-measurable. It follows that for any $\varepsilon>0$, we have a L-measurable set $$ E'_{\varepsilon}=(E\setminus E_{\varepsilon})\cup E_{\varepsilon}\supset E $$ with $m^*(E'_{\varepsilon}\setminus E)=m^*(E_{\varepsilon}\setminus E)\leq\sum_{n=1}^{\infty} \varepsilon/2^n=\varepsilon$.

Now consider $E'_{1/n}$ and let $E'=\cap_{n=1}^{\infty}E'_{1/n}$. Then $E'$ is L-measurable and $E'\supset E$. We also have $$ m^*(E'\setminus E)\leq m^*(E'_{1/n}\setminus E)\leq 1/n $$ for all $n$. Thus $m^*(E'\setminus E)=0$ and hence $E'\setminus E$ is L-measurable. It follows that $E=E'\setminus(E'\setminus E)$ is L-measurable. Q.E.D

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Here's an alternative approach, working more or less directly from the definitions. Let an arbitrary $\varepsilon>0$ be given; I want to produce an open $U\supseteq E$ with $m^*(U-E)\leq\varepsilon$. Apply the given hypothesis with $\varepsilon/2$ in place of $\varepsilon$ to get a measurable set $E'$ such that $m^*(E\triangle E')\leq\varepsilon/2$. Since $E'$ is measurable, there is an open $V\supseteq E'$ with $m^*(V-E')\leq\varepsilon/2$. Also, by definition of $m^*$, there is an open $W\supseteq E\triangle E'$ with $m(W)\leq\varepsilon/2$. Let $U=V\cup W$. Then $U$, being the union of two open sets, is open. Also, $U\supseteq E$; indeed, for any $x\in E$, if $x\in E'$ then $x\in V$, and otherwise $x\in E\triangle E'\subseteq W$, so in either case $x\in U$. It remains to prove that $m^*(U-E)\leq\varepsilon$, and for this purpose it suffices to show that $U-E\subseteq(V-E')\cup W$, because both $V-E'$ and $W$ have (outer) measure $\leq\varepsilon/2$ and outer measure is subadditive and monotone. So consider any $x\in U-E$; I want to show that $x$ is in (at least one of) $V-E'$ or $W$. By definition of $U$, $x$ is in $V$ or in $W$, and if it's in $W$ then we're done, so from now on suppose $x\in V$. If $x\notin E'$ then $x\in V-E'$ so we're again done. There remains only the case that $x\in E'$. But $x\notin E$, so $x\in E\triangle E'\subseteq W$, and we're again done.

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"..by definition of $m^*$, there is an open $W\supset E\Delta E'$ with $m(W)\leq \varepsilon/2$"? –  Jack Jan 5 '13 at 0:12
    
Do you mean the outer regularity of $m^*$? –  Jack Jan 5 '13 at 0:15
    
@Jack: I was using the definition of $m^*$ as the infimum of the measures of open supersets. –  Andreas Blass Feb 13 '13 at 21:06
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