In measure theory, given sets $A_1,A_2,\ldots$, we define $\liminf A_n=\bigcup_{k=1}^\infty\left(\bigcap_{n\geq k}A_n\right)$ and $\limsup A_n=\bigcap_{k=1}^\infty\left(\bigcup_{n\geq k}A_n\right)$.

What is the relation to the normal liminf/limsup for sequences? $\liminf{a_n} = \lim_{n\rightarrow\infty}\inf(a_n,a_{n+1},\ldots)$ and $\limsup{a_n} = \lim_{n\rightarrow\infty}\sup(a_n,a_{n+1},\ldots)$. How can I remember which one is the union of intersections, and which one is the intersection of unions?

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    $\begingroup$ To answer the last question, think of the $\liminf$ of sets as being the set which is contained in all but finitely many sets of the sequence (of sets). The $\limsup$ is the set of points which appear infinitely often (though they might miss infinitely many). $\endgroup$ – Clayton Sep 17 '13 at 15:54
  • $\begingroup$ @Clayton Yup, I know that interpretation. It's still hard for me to remember whether "infinitely often" and "almost always" correspond to $\liminf$ or $\limsup$. Just wondering if there's any other way to remember it? $\endgroup$ – PJ Miller Sep 17 '13 at 15:56
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    $\begingroup$ PJ, then appearing infinitely often means it will always appear in a union indexed for $k\geq n$. Appearing almost always means that it is always in the intersection indexed for $k\geq n$. This takes care of the inner part of the union/intersection combination. The outer part is just the opposite of the inner part. $\endgroup$ – Clayton Sep 17 '13 at 15:58
  • $\begingroup$ @Clayton Sure, thanks for your help! $\endgroup$ – PJ Miller Sep 17 '13 at 16:00

We can "identify" each set with its characteristic function

$$\chi_A(x) = \begin{cases}1 &, x \in A\\ 0 &, x \notin A.\end{cases}$$

Then we have

$$\chi_{\liminf A_n}(x) = \liminf \chi_{A_n}(x); \quad \chi_{\limsup A_n}(x) = \limsup \chi_{A_n}(x)$$

for all $x\in X$. The characteristic function of the limes inferior/superior of the sequence of sets is the pointwise limes inferior/superior of the characteristic functions of the sets in the sequence.

  • $\begingroup$ Perfect! Didn't realize this connection before. $\endgroup$ – PJ Miller Sep 17 '13 at 15:58

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