# Equivalence of Lebesgue Measurablity

Hello Mathematics 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.

Definitions

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.

Problem Statement

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.

This is what I have so far:

Let $T$ be a $G_{\delta}$ set. By definition, we have $T=\displaystyle \bigcap_{n=1}^{\infty} T_n$. Using De Morgan's Laws, $E \setminus \bigcap_{n=1}^{\infty} T_n$ = $\displaystyle \bigcup_{n=1}^{\infty} \left( E \setminus T_n \right)$ which is a closed set since its complement was open, by definition. Therefore, $\displaystyle \bigcup_{n=1}^{\infty} \left( E \setminus T_n \right)$ 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.

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.

• Have you established yet that if a set is measurable, then so is its complement? – Omnomnomnom Sep 15 '14 at 16:33
• $E\setminus T_n$ are not necessarily closed, they are closed in $E$, but might not be in $\mathbb{R}^d$. I will show you a general idea from 1 to 3. First assume that $E$ has finite measure. Given that $E$ is measurable, for each $n\in \mathbb{N}$, there exists an open set $U_n \supset E$ such that $$m(U_n \setminus E) \leq \frac{1}{n}.$$ Now define your $G_\delta$ set to be $\cap_{n=1}^\infty U_n$. What can you say about this set? – Xiao Sep 15 '14 at 16:51
• Hello, and thanks for the constructive feedback. We have established that the compliment of a measurable set is also measurable. – Jamil_V Sep 15 '14 at 20:15
• As for the idea from 1 to 3, if I define my $G_{\delta}$ set as $\bigcap_{n=1}^{\infty}U_n$, then won't it's measure also be less than or equal to $\frac{1}{n}$? If this is the case, then what happens if I let $\epsilon = \frac{1}{n}$? – Jamil_V Sep 15 '14 at 20:28
• If the quantity $m(\cap_{n=1}^\infty U_n \setminus E)$ is less than or equal to $\frac{1}{k}$ for each $k\in \mathbb{N}$, see below $$m(\cap_{n=1}^\infty U_n \setminus E) \leq m(\cap_{n=1}^k U_n \setminus E) \leq m(U_k \setminus E) \leq \frac{1}{k}$$ what can you say about $m(\cap_{n=1}^\infty U_n \setminus E)$? – Xiao Sep 15 '14 at 22:21