Equivalence of Lebesgue Measurability

[jamilmalik]

Homework Statement

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.

Homework 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.

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

 

[mighty2000]

Hello there! It looks like you are having some difficulty with this problem. I am happy to help you work through this and provide some guidance.

First, let's review the definitions provided. A set E is said to be Lebesgue measurable if, for every ε>0, there exists an open set U⊂Rd containing E such that m*(U∖E)≤ε. This definition essentially means that E can be "approximated" by open sets with arbitrarily small measure.

Next, we have the definitions of Gδ and Fσ sets. A Gδ set is a countable intersection of open sets, while an Fσ set is a countable union of closed sets. These definitions are important because they allow us to express sets in terms of "simpler" sets (open and closed sets are often easier to work with).

Now, let's look at the three statements that we need to show are equivalent.

1. E is Lebesgue measurable.
2. E is a Gδ set with a null set removed.
3. E is the union of an Fσ set and a null set.

To show that these statements are equivalent, we need to show that each one implies the other two. Let's start with statement 1.

Assume that E is Lebesgue measurable. This means that for every ε>0, there exists an open set U⊂Rd containing E such that m*(U∖E)≤ε. We can also write this as m*(E∪(U∖E))≤ε. Now, let's define a set F=E∪(U∖E). F is an Fσ set (since it is the union of E and an open set, which is a countable union of closed sets). However, we need to show that F is also a null set. This means that m*(F)=0.

To show this, we can use the fact that m*(E∪F)=m*(E)+m*(F). Since m*(E∪F)≤ε and m*(E)=0 (since E is measurable), we have m*(F)≤ε. Since ε was arbitrary, this means that m*(F)=0, and therefore F is a null set. This shows that statement 1 implies statement 3.

Now, let's show that statement 3