This problem was on my in-class final for a measure theory course I took in the fall, and now I am studying for my qualifying exam so I am trying to figure this one out:
Suppose $\{E_n\}_{n=1}^{\infty}$ is a sequence of measurable sets in $X$. Suppose $$A= \{x\in X :x\in E_n\hspace{2.5mm} \text{for infinitely many}\hspace{2.5mm} n\in \mathbb{N}\} $$ and $$B= \{x\in X :x\in E_n\hspace{2.5mm} \text{for all but finitely many}\hspace{2.5mm} n\in \mathbb{N}\} .$$
Prove $A=B$ when the sets $E_n$ are nested increasing ($E_n\subset E_{n+1}$) as well as when they are nested decreasing ($E_{n+1}\subset E_n$).
I know that we can rewrite $$A=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}E_k$$ and $$B=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty} E_k.$$ I have tried a couple different approaches to this problem and I keep getting stuck pretty early on. Any help would be appreciated.
Best Answer
$B\subset A$ regardless of the assumptions on $E_n$, as this is just a basic property that $\liminf E_n\subset\limsup E_n$. So we just need to show the reverse inclusion holds based on these assumptions.
Suppose the sets $E_n$ are nested increasing, $E_{n}\subset E_{n+1}$. Let $x\in A$. Let $A_x=\{n\in\mathbb{N} : x\in E_n\}$. $A_x\subset\mathbb{N}$ is nonempty and infinite. There exists a least element $n_0\in A_x$ by elementary properties of the natural numbers. Therefore $x\in E_{n_0}\Rightarrow x\in E_n \forall n\geq n_0$. Then $x\not\in \bigcup_{1}^{n_0-1} E_i$ which is finite, so $x\in B\Rightarrow A\subset B$.
Suppose the set $E_n$ are nested decreasing, $E_{n+1}\subset E_{n}$. Let $x\in A=\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$. For each $n$, $\bigcup_{k=n}^{\infty} E_k=E_n$ by the nested decreasing, so $ A=\bigcap_{n=1}^{\infty} E_n$. In particular, $x\in A$ implies $x\in E_n$ for every $n$, implying that $x$ is in all but finitely many $E_n$ (in fact, $0$), so $x\in B$ implies $A\subset B$.
(Alternatively, $A=\bigcap_{n=1}^{\infty} E_n\subset \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} E_k=B$)
I used two different techniques. In reality, the technique for the second one works for both, and you can directly show equivalence by using the intersection/union definition. I just did the first case differently because that's the first way I thought of doing it.
This proof requires that neither $A,B$ are empty. By dealing just with the set definitions themselves, you don't have to worry about this extra step.