(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

Let E be a subset ofRand suppose the Lebesgue outer measure m*(E) is finite. Prove that the following two statements are equivalent:

(1) Given e > 0, there is an open set A containing E with m*(A\E) < e.

(2) Given e > 0, there is a finite union V of open intervals such that m*(V\E U E\V) < e.

Relevant equations

Here's a statement that may come in handy: For any subset X ofRand any e > 0, there is an open set Y such that Y contains X and m*(Y) <= m*(X) + e.

The attempt at a solution

I haven't even bothered trying to prove that (2) implies (1) as I'm stuck on the proof that (1) implies (2). Here's what I have so far: Let e > 0. By (1), there is an A containing E with m*(A\E) < e/2. A is an open set, so it is the union of countably many open intervals. If "countably many" is finite, then we're done. Otherwise, start by picking an open interval from A that intersects E. Call it V. We have that m*(V\E) <= m*(A\E) < e/2. I need to prove that m*(E\V) < e/2. If I can't, then I can make V larger by adding to it another open interval from A that intersects E. I imagine that after doing this finitely many times, V will be large enough so that m*(E\V) < e/2. The problem is that I don't have any way of calculating m*(E/V). Any tips?

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