$\limsup = \liminf$ of sequence of Sets

Question

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.

Answer 1

$B\subset A$ 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.

Answer 2

Here is another way to explain this purely in terms of the "words" definition you gave for $A$ and $B$. This is not rigorous, but might help you see what others are posting in rigorous math-speak. Again, I'm assuming that $B \subset A$ is obvious so you don't need to show this. You want to show $A \subset B$. Start by observing that if $x \in A$, then it is in $E_n$ for a particular finite $n$, call it $N$.

If $E_n$ is an increasing sequence, then it is in all $E_n$'s after $E_N$. Therefore at most it is missing from the $E_n$ for $1 \leq n < N$, which is a finite number, so it has to be in $B$.

If $E_n$ is a decreasing sequence, then the same must be true. If not, then think about what must happen: it must not only jump out of the sequence at some point (i.e. $x \notin E_M$ for some $M > N$), but it must also jump back in because of the definition of $A$. But if $E_n$ is decreasing, then once the point leaves the sequence it can never come back in. Makes sense?