Equivalent definitions for Lebesgue measurable set

Question

Given $A \subseteq \mathbb{R}$ I am trying to show that the following are equivalent:

a) $A$ is Lebesgue measurable

b) There exists some $F_\sigma$ set $F$ and a set $Z$ with null measure such that $A = F \cup Z$ where every closed set in $F$ is included in $A$.

c) There is some $G_\delta$ set $G$ and $Z$ with null measure such that $A$ is included in every set of $G$ and $A = G_\delta \setminus Z$.

I've been able to prove that $a) \Rightarrow b)$ using the property that states that if $A$ is Lebesgue measurable then for every $\epsilon > 0$ there exists a closed set $F \subseteq A$ such that $\mu(A \setminus F) < \epsilon $.

Now $b) \Rightarrow c)$ shouldn't be hard. For now I have that $A = (\bigcup_{n \in \mathbb{N}} F_n) \cup Z$ where $F_n \subseteq A$ are closed sets and $Z$ is a set with null measure. Taking complement we get $\mathbb{R} \setminus A = (\bigcap_{n \in \mathbb{N}} \mathbb{R} \setminus F_n) \setminus Z$. How can I proceed from here to finish up that $A = (\bigcap_{n \in \mathbb{N}} G_n) \setminus Z$?

Does this answer your question? Inner regularity of Lebesgue measurable sets $$$$ Proof of Outer Regularity of Lebesgue Measure on $\Bbb R$ — PinkyWay, Jun 24 '23 at 20:57
Hint: for every $n$ there exists a closed set $F_n \subseteq A$ with $\mu(A \setminus F) < 1/n$. — Nate Eldredge, Jun 24 '23 at 21:05
If I take $\forall n \in \mathbb{N}$ the closed sets $F_n \subseteq A$ with $\mu(A \setminus F_n) < 1/n$ then $Z = \bigcap F_n \ \subseteq F_n$ and $\mu(Z \setminus A) = 0$ if I am not mistaken,. Is that what you were hinting at @Nate Eldredge? — Peter Sampodiras, Jun 24 '23 at 21:16
But is it true that $A = (\bigcup F_n ) \cup Z$? Is that the $F_\sigma$ I need? @Nate Eldredge — Peter Sampodiras, Jun 24 '23 at 21:24
Your definitions have gotten a bit garbled, I think. You seem to be using $F_\sigma$ to denote both the class of all $F_\sigma$ sets (i.e. all sets that can be written as a countable union of closed sets), as well as some particular set of that kind. I think the statement you are supposed to prove could be written more clearly as: there exists an $F_\sigma$ set $F$ and a null set $Z$ such that $A = F \cup Z$. — Nate Eldredge, Jun 24 '23 at 21:31
And then the argument goes: choose $F_n \subseteq A$ with $\mu(A \setminus F_n) < 1/n$. Let $F = \bigcup_n F_n$, so that $F$ is $F_\sigma$, and let $Z = A \setminus F$. By construction we have $A = F \cup Z$. Now verify that $\mu(Z)=0$. — Nate Eldredge, Jun 24 '23 at 21:32
Yes, you are right, I'll change the formulation of the question to make it a bit clearer. Thanks! — Peter Sampodiras, Jun 24 '23 at 22:11

Equivalent definitions for Lebesgue measurable set

0 Answers0