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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?


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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
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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