I've been trying to think about this problem for awhile now:

Let $\{f_k\}_{k=1}^{\infty}$ be a sequence of measurable functions. Prove that the functions f,g defined by:

$f(x) = \limsup_{k \rightarrow \infty} f_k(x)$

$g(x) = \liminf_{k \rightarrow \infty} f_k(x)$

are measurable.

Use this to prove that if $f_k \rightarrow f$ a.e. on domain $E$, then $f$ is measurable.

As far as ideas for proving that $f,g$ are measurable, I know that I can express each as a combintion of the sup and inf function, which are both measurable, but then a composition of measurable functions isn't guarentee'd to be measurable.. But it is an idea.

As far as proving that $f_k \rightarrow f$ a.e. implies $f$ is measurable, it seems reasonable enough, and I'm pretty sure this result was one of the motivating reasons for inventing measure theory in the first place, but I'm having problem getting a proof on paper. Comments, insights and proofs appreciated!! Thanks all!

  • 1
    $\begingroup$ Can you prove that the functions $\sup_kf_k(x)$ and $\inf_kf_k(x)$ is measurable? Then, can you find a definition of $\limsup f_k$ and $\liminf f_k$ that relates these two? $\endgroup$ Commented Sep 25, 2019 at 18:30

1 Answer 1


Prove that $$\{x: \limsup_nf_n(x)>a\}=\bigcap_{n=1}^{\infty}\bigcup_{k \geq n}^{\infty}\{x:f_k(x)>a\}$$ $$\{x: \liminf_nf_n(x)>a\}=\bigcup_{n=1}^{\infty}\bigcap_{k \geq n}^{\infty}\{x:f_k(x)>a\}$$

So you will have the conclusion about the measurability of limsup and liminf of the functions.

Also note that $f(x)=\limsup_nf_n(x)$ a.e so $f$ is measurable.


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