# Measure theory limsup, liminf

• Feb 24th 2014, 11:06 PM
sawleha
Measure theory limsup, liminf
I've just started measure theory and I'm really struggling with what's supposed to be easy proofs. Any help would be appreciated.

1.1) Let $(E_n)_{n\epsilon N}$ be a sequence of subsets of X. Prove that $\limsup E_n =\bigcap _{k=1}\bigcup_{n\geq k} E_n$

1.2) Let $(E_n)_{n\epsilon N}$ be a sequence of subsets of X. Prove that $\liminf E_n =\bigcup _{k=1}\bigcap_{n\geq k} E_n$

I attempted to do this using the definition of limsup and liminf (for limsup. x belongs to limsup if it belongs to infinitely many E_n and y belongs to liminf if it belongs to all but finitely many E_n) but my professor said to rather do it using set inclusion that is if x belongs to limsupE_n then x belongs to $\bigcap _{k=1}\bigcup_{n\geq k} E_n$ and if y belongs to $\bigcap _{k=1}\bigcup_{n\geq k} E_n$ then y belongs to limsup E_n. I started by assuming $(x\epsilon limsup E_n$ and $y\epsilon \bigcap _{k=1}\bigcup_{n\geq k} E_n$ where do I go from here?
• Feb 25th 2014, 04:55 AM
emakarov
Re: Measure theory limsup, liminf
I recommend first grasping the intuitive idea behind $\displaystyle\bigcap_{k=1}\bigcup_{n\ge k}E_n$ and $\displaystyle\bigcup_{k=1}\bigcap_{n\ge k}E_n$. When you have a clear picture in your head, you'll be able to write a formal proof.

For example, $\displaystyle\bigcap_{n\ge k}E_n$ consists of elements belonging to all of $E_k$, $E_{k+1}$, ... . Eeach such element belongs to all but finitely many $E_n$'s. But elements of $\displaystyle\bigcap_{n\ge k+1}E_n$ and $\displaystyle\bigcap_{n\ge 2k}E_n$ also have this property. If you want to collect all elements that belong to every $E_n$ from some point, i.e., to form $\liminf E_n$, you need to take the union over $k$: $\displaystyle\bigcup_{k=1}\bigcap_{n\ge k}E_n$.

Similarly, $\displaystyle\bigcup_{n\ge k}E_n$ consists of elements that belong to some of $E_k$, $E_{k+1}$, ... ; that is, they occur in one of the sets after index $k$. If we take the intersection over $k$, we'll get the set of elements that belong to one of the sets starting from any point. I.e., they belong to one of the sets after $E_5$, they belong to one of the sets after $E_{100}$ and so on. That is, the resulting set has elements that belong to infinitely many $E_n$'s, i.e., $\limsup E_n$.
• Mar 2nd 2014, 11:23 PM
sawleha
Re: Measure theory limsup, liminf
I think I understand what you're saying. If http://latex.codecogs.com/png.latex?..._{n\geq k} E_n then y belongs to ( $(E_k or E_{k+1} or E_{k+2}...) and (E_{k+1} or E_{k+2} or E_{k+3}...)$ so y belongs to one of the sets after k, by taking the intersection of these sets we find that y is in infinitely many sets and by definition $y \epsilon limsupE_n$ conversely starting with an element of limsup, since it is in infinitely many sets it must be in some set $E_n$ for $n \geq k$ and so it is in the union, but it must be in infinitely many sets so it's in the intersection of the union and so by set containment, the two are equal? I think I understand it, but this looks intuitive rather than formal?
• Mar 2nd 2014, 11:48 PM
ThePerfectHacker
Re: Measure theory limsup, liminf
It is useful to think of $\cap$ symbol as the word "for all" and $\cup$ symbol as the word "there exists".

When you say,
$$x\in \bigcup_{n\geq 1} \bigcap_{k\geq n} A_k$$
You are saying,
There is an $n\geq 1$ such that for all $k\geq n$, $x\in A_k$. This can be restated by saying that $x$ is therefore in almost all $A_k$ (because from some point on $x$ belongs to all of them).

When you say,
$$x\in \bigcap_{n\geq 1}\bigcup_{k\geq n}A_k$$
You are saying,
For every $n\geq 1$ there is an $k\geq n$ so that $x\in A_k$. In other words, $x$ must be in infinitely many of the sets as no matter how far you go into the sequence of sets at some point $x$ will belong to some set.
• Mar 3rd 2014, 12:01 AM
sawleha
Re: Measure theory limsup, liminf
Wow. Two phrases bring so much context into it. Thank you.