Proving Equivalence of (1) and (2): On Measurability

[e(ho0n3]

Homework Statement

Let E be a subset of R and 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 of R and 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?

 

[e(ho0n3]

I have figured out the following: A is the union of countably many disjoint open intervals, say I_1, I_2, ... Let V_i = E \ (I_1 U ... U I_i). We have that m*(V_1) is finite and that V_1 contains V_2 contains V_3 etc. Hence, m*(V_1) >= m*(V_2) >= ..., i.e. (m*(V_i)) is a bounded sequence of positive integers, so it converges to inf {m*(V_i)}. Now all I need to do is show that said inf is in fact 0. This, I have not been able to do. Any tips?