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My instructor gave an outline for how to proof the following

For $E \subset \mathbb{R}$ , There is a $G_{\delta}$ set $U \supseteq E$ such that $m^*(E) = m^*(U)$.

I have a solution, but I'm unsure about some of the details.

My solution:

Case 1. If $m^*(E) = \infty$ then we can choose our $G_{\delta}$ set $U = \displaystyle \bigcap_{n=1}^\infty (- \infty, \infty) = \mathbb{R}$.

Case 2. Suppose $m^*(E) < \infty$. Then there exists a cover $\bigcup_{n=1}^\infty I_n^j \supseteq E$ such that

$m^*(E) +\displaystyle \frac{1}{j} \geq \sum_{n=1}^\infty L(I_n^j)$. Define our $G_{\delta}$ set $U = \bigcap_{j=1}^\infty ( \bigcup_{n=1}^\infty I_n^j)$. Then because $E \subseteq \bigcup_{n=1}^\infty I_n^j$, we have $E \subset U$. Then by monotonicity we have $m^*(E) \leq m^*(U)$. By the definition of $U$, we have that $U \subseteq \bigcup_{n=1}^\infty I_n^j$. This means that

$m^*(E) + \displaystyle \frac{1}{j} \geq \sum_{n=1}^\infty L(I_n^j) \geq m^*(U)$. So if we let $j$ go to infinity we have $m^*(E) \geq m^*(U)$.

So we have shown that $m^*(E) \leq m^*(U)$ and $m^*(E) \geq m^*(U)$. Thus we must have $m^*(E) = m^*(U)$.

What I'm confused about is why we need two variables on our interval. I'm confused about how $j$ and $n$ are related and how it affects the proof.

Thanks.

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    Look at it this way: For any $n$ there is an open $U_n$ such that $E \subset U_n$ and $ m^* E \le m^* U_n \le m^* E + {1\over n}$. Let $V = \cap_n U_n$. – 2017-02-05
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    For Case 1 you do not need the intersection; you are not intersecting any sets there. – 2017-02-05
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    For case 2 you only claim there exists *a* cover of n-intervals meeting the stated criteria (assumedly for some $j$, perhaps in $\mathbb{N}?$) and then intersect over an infinite amount of different n-intervals; make sure you establish that ALL these covers exist, not just "a" cover. Make sure to state E is in every one of these covers. $n$ is needed to denumerate all intervals in each *specific* cover. $j$ denumerates the *covers themselves,* with $\frac{1}{j}$ the bound for how much the intervals in that jth cover "overshoot" the measure of $E$. – 2017-02-05

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