Equivalence of Lebesgue Measurability

1. Sep 17, 2014

jamilmalik

1. The problem statement, all variables and given/known data

Hello Physics Forums Community. I am having some difficulties with the following problem dealing with Lebesgue Measure and its equivalent interpretation. I will first include the definitions which I am using and then the problem statement, they are coming straight from Terrence Tao's Introduction to Measure Theory book which is available for free online (Link provided at the end of this post).

Let $E \subset \mathbb{R^d}$. Show that the following are equivalent:

1. $E$ is Lebesgue measurable.
2. $E$ is a $G_{\delta}$ set with a null set removed.
3. $E$ is the union of a $F_{\sigma}$ set and a null set.

2. Relevant equations

A set $E \subset \mathbb{R^d}$ is said to be Lebesgue measurable if, for every $\epsilon >0$, there exists an open set $U \subset \mathbb{R^d}$ containing $E$ such that $m^*(U \setminus E) \leq \epsilon$.

Define a $G_{\delta}$ set to be a countable intersection $\displaystyle \bigcap_{n=1}^{\infty} U_n$ of open sets.

Define a $F_{\sigma}$ set to be a countable union $\displaystyle \bigcup_{n=1}^{\infty}F_n$ of closed sets.

3. The attempt at a solution

This is what I have so far:

Assume that $E$ has finite measure. Let $U$ be a $G_{\delta}$ set. By definition, we have $U=\displaystyle \bigcap_{n=1}^{\infty} U_n$. Using De Morgan's Laws, $E \setminus \bigcap_{n=1}^{\infty} U_n$ = $\displaystyle \bigcup_{n=1}^{\infty} \left( E \setminus U_n \right)$ which is a closed set since its complement was open, by definition. Therefore, $\displaystyle \bigcup_{n=1}^{\infty} \left( E \setminus U_n \right)$ is a countable union of closed sets which is the definition of a $F_{\sigma}$ set. Although this shows a relationship between the $G_{\delta}$ sets and the $F_{\sigma}$ sets, I don't know how to include the condition of the null set removed or added as well as connect to the Lebesgue measure.

This first approach didn't seem so promising as I do not think that it was leading to anything useful. Here is another approach which I believe has a more promising direction, but I am getting stuck in completing the proof.

Given that $E$ is measurable, for any $n \in \mathbb{N}$, then there exists an open set $U_n \supset E$ such that $m(U_n \setminus E) \leq \frac{1}{n}$.
If the quantity $m(\cap_{n=1}^{\infty}U_n \setminus E)$ is less than or equal to $\frac{1}{k}$ for each $k \in \mathbb{N}$, then

$m(\cap_{n=1}^\infty U_n \setminus E) \leq m(\cap_{n=1}^k U_n \setminus E) \leq m(U_k \setminus E) \leq \frac{1}{k}$. If I let $\epsilon = \frac{1}{k}$, then isn't this equivalent to the definition of Lebesgue measurability above? In other words, if I let by $G_{\delta}$ set to be $\cap_{n=1}^{\infty}U_n$, then doesn't the implication $1 \Rightarrow 2$ follow?
As for the other implications, I do not know how to start.

I would greatly appreciate any help that I can receive on this problem since I cannot seem to proceed in the right direction. Thanks in advance.

Here is the link for the free version of the textbook:
http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf

Last edited: Sep 17, 2014
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