I have a problem with the lim sup definition when it is applied to a sequence of functions.

Consider the sets $(A_{n})_{n\in\mathbb{N}}$. They are all pairwise disjoint.

Then we have a sequence of functions given:

$f_{n} = 1_{A_{n}}$

My problem is to compute $\limsup\limits_{n\rightarrow\infty} f_{n}$.

Based on my textbook I have tried to compute it using the following definition:

$\limsup\limits_{n\rightarrow\infty} f_{n} = \lim_{k\to\infty} \sup_{j\geq k}{1_{A_{j}}}(x) $

I decide to divide the problem into two subproblems:

  • $x$ is in one of the sets (let us denoted it $A_{t}$ where $t\in\mathbb{N}$)
  • $x$ is not contained in any of the sets.

In the first case I have deduced that when $k$ goes to infinity, it will get bigger than $t$. I then calculate the above as:

$\limsup\limits_{n\rightarrow\infty} f_{n} = \lim_{k\to\infty} \sup_{j\geq k} 1_{A_{j}}(x) = 0 $

In the last case I also get $0$ - because $f_n$ will always return $0$.

Is this correct? Do I "get" the definition? I'm not sure, because the book mentions something about it should be understood "pointwise". And I don't exactly see how it should be understood "pointwise". I think I am doing it pointwise, but I'm not sure..

  • $\begingroup$ Please replace $\lim_{n\to+\infty}\sup$ by $\limsup\limits_{n\to\infty}$, thrice. $\endgroup$ – Did Sep 22 '12 at 13:48
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    $\begingroup$ Yes, you are doing this poitwise as you fixed $x$, and saw what happens for this $x$. $\endgroup$ – Davide Giraudo Sep 22 '12 at 15:28
  • $\begingroup$ What you have so far is correct, but you are missing a case: $x$ could be contained in more than one of the sets $A_n$. The key distinction is whether it is contained in finitely or infinitely many of them. $\endgroup$ – Nate Eldredge Sep 22 '12 at 16:09
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    $\begingroup$ No this is not a case, because the sets $A_{n}$ are pairwise disjoint :) $\endgroup$ – whoisitnow Sep 22 '12 at 17:17
  • $\begingroup$ @whoisitnow: Oh, I missed that :) More generally, you might like to prove that for arbitrary sets $A_n$, $\limsup 1_{A_n} = 1_A$, where $A = \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty A_n$. $\endgroup$ – Nate Eldredge Sep 23 '12 at 13:47

As you fixed $x$, the proof has been done pointwise, as expect. Maybe we just need to say that there are only the two cases you are dealing with as the $A_n$ are pairwise disjoint.


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