Let $(X,\Sigma,\mu)$ be a measure space and let $\{E_n\}_{n=1}^\infty$ be a sequence of sets in $\Sigma$. I try to show that $$\lim_{m\to \infty} \mu(\cap_{n=m}^\infty E_n)\leq \liminf \mu(E_n)$$ by using the Fatou's lemma.

Attempt: Let $f_n=\chi_{E_n}$. By Fatou's lemma, we have $\liminf\int f_n \ge\int\liminf f_n$. Clearly, $\liminf\int f_n=\liminf \mu(E_n)$. So it remains to show that $\int \liminf f_n \ge \lim_{m\to \infty}\mu(\cap_{n=m}^\infty E_n)$

I know this link has the same question but I want to conclude my attempt. Thanks!


1 Answer 1


Hint: Since $f_n$ only takes the values $0$ and $1$, the same is true of $\liminf f_n$. Hence $\liminf f_n$ is itself the characteristic function of some set. Try to find out which set.

  • $\begingroup$ $\int \liminf f_n=lim_{m\to\infty}\int \inf_{n\ge m} \chi_{E_n}$. But here I cannot define $\inf_{n\ge m}\chi_{E_n}$. So I cannot take integral of it. $\endgroup$ Nov 2, 2014 at 22:11
  • $\begingroup$ For which $x$ is $\inf_{n\ge m}\chi_{E_n}(x)=1$? For which $x$ is it zero? (Do you see why it has no other values than those two?) Express your answers in terms like “for all $n\ge m$ …” or “for some $n\ge m$ …”, then interpret that as unions or intersections. $\endgroup$ Nov 3, 2014 at 0:05

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