Let $(X,\alpha,\mu)$ be a measure space. Let $(E_n)$ be a sequence in $X$.

Prove $\limsup\mu(E_n)\leq\mu(\limsup E_n)$ when $\mu(\bigcup E_n)<\infty$.

Show that this inequality may fail if $\mu(\bigcup E_n)=\infty$.


I was thinking of showing the reverse - $\mu(\limsup E_n)\geq\limsup\mu(E_n)$ because I know $\limsup E_n=\bigcap _{n=1}^\infty\bigcup_{m=n}^\infty E_n$. I know the measure of a union is equal to the union of the measures if the sequence is disjoint (which I am not guaranteed to have here). Besides I would have to deal with the intersection first.

I also have to prove a similar inequality for $\liminf$, but I figure I should be able to figure out the other if I can figure out one.

Please let me know what ideas you have. Thank you!


1 Answer 1


Let $A = \cup E_n$. By Fatou's lemma,

$$\int_A \liminf 1_{A\setminus E_n} \, d\mu \le \liminf \int_X 1_{A\setminus E_n}\, d\mu.$$


$$\int_A \liminf (1 - 1_{E_n})\, d\mu \le \liminf \int_A(1 - 1_{E_n})\, d\mu$$


$$\mu(A) - \int_A \limsup 1_{E_n}\, d\mu \le \mu(A) - \limsup \int_A 1_{E_n}\, d\mu.$$

Since $\limsup 1_{E_n} = 1_{\limsup E_n}$, $\int_A 1_{E_n}\, d\mu = \mu(E_n)$, and $\int_A 1_{\limsup E_n}\, d\mu = \mu(\limsup E_n)$, we have

$$\mu(A) - \mu(\limsup E_n) \le \mu(A) - \limsup \mu(E_n).$$

Since $\mu(A) < \infty$, we deduce

$$\limsup \mu(E_n) \le \mu(\limsup E_n).$$

  • $\begingroup$ How I am allowed to just let $X=\bigcup E_n$? Also, I am unfamiliar with the "1" notation - is it some sort of characteristic equation? $\endgroup$
    – fullyhip
    Mar 1, 2015 at 1:01
  • $\begingroup$ @fullyhip Saying $X = \bigcup E_{n}$ is just giving the union a name (the name $X$) so that we don't have to keep writing $\bigcup E_{n}$ over and over. Also, since $E_{n}$ is a set for each $n$, of course you can take the union of all of these sets. So it makes sense to talk about the union over all $n$ of $E_{n}$. $\endgroup$
    – layman
    Mar 1, 2015 at 1:02
  • $\begingroup$ If $A$ is a set, then $1_A(x)$ equals $1$ if $x \in A$ and $0$ if $x \notin A$. $\endgroup$
    – kobe
    Mar 1, 2015 at 1:02
  • $\begingroup$ @kobe What is $1 - 1_{E_{n}}$? Is that the constant function $1$ minus $1_{E_{n}}$? $\endgroup$
    – layman
    Mar 1, 2015 at 1:03
  • 1
    $\begingroup$ Excellent...GRAZIE MILLE KOBE! $\endgroup$
    – fullyhip
    Mar 1, 2015 at 1:16

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