Measurability of the set of elements who belong to a infinite amount of subsets in a sequence

Question

I've been struggling to prove the following statement:

Let (X,$\mathcal{M}$,$\mu$) a finite measure space and let $(A_n)_{n\in\mathbb{N}}$ a sequence of measurable sets in X. Now consider $M$ the set of elements from X who belong to a infinite amount of the subsets $A_n$. Prove $M$ is measurable and $$\inf_{n\in\mathbb{N}} \mu(A_n) \leq \mu(M)$$

I have checked for a solution here but i do not understand neither the solution given (infinite does not mean "all") nor the approximation from the person who asked. Thanks in advance.

Answer 1

$x\in A_n$ for infinitely many $n$ is equivalent to saying that for any $m\in\mathbb{N}$ there exists a $n\in\mathbb{N}$ such that $n\geq m$ and $x\in A_n$ which is further equivalent to the fact that $x\in\cup_{n=m}^{\infty}\,A_n$ for all $m\in\mathbb{N}$. So we get that your required set of points is $M=\cap_{m=1}^{\infty}\,\cup_{n=m}^{\infty}\,A_n$. Since, countable union/intersection of measurable sets is measurable, $M$ is measurable. Since, $\mu$ is a finite measure we have, $\mu(M)=\lim_{m\to\infty}\mu(\cup_{n=m}^{\infty}\,A_n)$. Again observe that $\mu(\cup_{n=m}^{\infty})\geq\mu(A_n)\forall n\geq m$ for each fixed $m$. Again, $\mu(A_n)\geq\inf\mu(A_n)$. Now take limit over $m$ to achieve the desired inequality.

Answer 2

Not an answer, just want to add something that could be nice to know. For a sequence of sets $(A_n)_{n\in\mathbb N}$, the set $$ A:=\bigcap_{n\geq 1}\bigcup_{m\geq n} A_n $$ is called 'limsup' of the sequence $(A_n)$, written $$ A=\limsup_{n\to\infty} A_n. $$ There is an analogous definition for the liminf. One well-know exercise in measure theory, which is basically equivalent to the one in your post, is to show that if a sequence of sets $(A_n)$ is contained in a $\sigma$-algebra $\mathcal F$, then $\limsup A_n$ and $\liminf A_n$ both belong to $\mathcal F$.

Check the wikipedia page.