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I am trying to show that the following statement is equivalent to proving that a set $E\subset\mathbb{R}$ is Lebesgue measurable:

Given $\epsilon>0$ there is an open set $O\supset E$ with $m^*(O\setminus E)<\epsilon$. ($m^*$ is the outer measure.)

The hint in my book states that we can find open sets $O_n$ and closed sets $F_n$ such that $m^*(\cap_{n\in\mathbb{N}} O_n\setminus E)=m^*(E\setminus \cup_{n\in\mathbb{N}} F_n)=0$. Thus $E$ is Borel upto a null set and so Lebesgue measurable. I can establish the existence of the requisite $O_n$ through previous theorems in the book for the converse, i.e. for Lebesgue measurability implies this statement, but am not sure how to conclude Lebesgue measurability from this. Can someone explain?

Thanks.

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Have you chosen the sets so that $O_n\supseteq E\supseteq F_n$ for each $n$? –  Brian M. Scott Oct 1 '12 at 8:24
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Such an $O_n$ exists by the statement. I am not sure how to establish the existence of $F_n$. So I can prove $m^*(\cap O_n\setminus E)=0$ but not the other equality. –  Shahab Oct 1 '12 at 12:21
    
What definition of measurable are you using? –  leo Oct 1 '12 at 22:49
    
A set E is Lebesgue measurable iff $\forall A\subset \mathbb{R}$, we have $m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$. Here $m^*$ is the outer measure defined $\forall A\subset \mathbb{R}$ so that $m^*(A)=inf\{\sum_{n\in\mathbb{N}}l(I_n):I_n$ are intervals, $A\subseteq\cup_{n\in\mathbb{N}}\}$. ($l(I_n)$ means length of the interval $I_n$.) –  Shahab Oct 2 '12 at 4:56

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