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Let $E\subset R^{d}$,show that the followig are equivalent:

(a)$E$ is Lebesgue measurable

(b)for every elementary set A, one has $m(A)=m^{*}(A\setminus E)+m^{*}(A\cap E)$, where $m^{*}$ is defined as the Lebesgue outer measure.

and our definition of Lebesgue measurable set is: Given $\epsilon>0$ there is an open set $O\supset E$ with $m^*(O\setminus E)<\epsilon$.

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Suppose $E\in\mathcal{M}$. Then for any $A\subseteq\Bbb R^n$ there exists $H\supseteq A$ such that $H$ is a $G_\delta$-set and $|A|_e=|H|$. Now, write $H=(H\cap E)\cup (H\cap E^C)$, a disjoint union of measurable sets, so $|A|_e=|H|=|E\cap E|+|H-E|$. Now, observe $A\cap E\subseteq H\cap E$ and $A\cap E^C\subseteq H\cap E^C$, so $|A\cap E|_e+|A\cap E^C|_e\leq|A|_e$. The reverse inequality follows from subadditivity of outer measure. Thus, $$ |A|_e=|A\cap E|_e+|A\cap E^C|_e. $$ Conversely, we want to show if $|A|_e=|A\cap E|_e+|A\cap E^C|_e$ for any $A\subseteq\Bbb R^n$, then $E$ is measurable. Suppose $|E|_e<\infty$. Then we can find a $G_\delta$-set $H\supseteq E$ with $|H|=|E|_e$. Write $H=(H\cap E)\cup (H\cap E^C)$. Then $H\cap E=E$ and we can apply the hypothesis with $A=H$ and see that $$ |E|_e=|H|=|H|_e=|E|_e+|H-E|_e. $$ Since $|E|_e<\infty$, we can subtract it from both sides, so $|H-E|_e=0$, thus $H-E$ is measurable and $|H-E|=0$. Define $Z=H-E$, then $E=H-Z$, therefore $E\in\mathcal{M}$.

In the case that $|E|_e=\infty$, let $B_k$ be the ball with center $\mathbf{0}$ and radius $k$. Then each $E_k=E\cap B_k$ has finite outer measure and $E=\bigcup_{k\in \Bbb N}E_k$. Now, let $H_k$ be a $G_\delta$ containing $E_k$ with $|H_k|=|E_k|_e$. By hypothesis, $$|H_k|=|H_k\cap E|_e+|H_k-E|_e\geq |E_k|_e+|H_k-E|_e.$$ Therefore, $|H_k-E|=0$. Let $H=\bigcup H_k$. Then $H$ is measurable, $E\subset H$, $H-E$ has measure zero, hence is measurable. Since $E=H-(H-E)$, we have that $E$ is measurable.

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  • $\begingroup$ but here your $H$ is $G_{\delta}$, not an elementary set, which is defined as the finite union of boxes. Is there a way to rectify this proof a little so as to make it right?@Clayton $\endgroup$ – Alex Jan 14 '13 at 5:24
  • $\begingroup$ For each $n\in \Bbb N$, there exists an open $H_n$ such that $E\subseteq H_n$ and $|H_n-E|<\frac{1}{n}$. Set $H=\bigcap H_n$. As for the finite union, any open set can be written as a countable union of nonoverlapping boxes, i.e., we can take finitely many of them and make the difference as small as we like. $\endgroup$ – Clayton Jan 14 '13 at 6:21
  • $\begingroup$ Oh, I see. You want to use the $\frac{\epsilon}{2^{n}}$ trick, is it? $\endgroup$ – Alex Jan 14 '13 at 15:05

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