2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.