# Proving that the $\limsup$ and $\liminf$ of a sequence of measurable functions are measurable

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!

• 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? Commented Sep 25, 2019 at 18:30

Prove that $$\{x: \limsup_nf_n(x)>a\}=\bigcap_{n=1}^{\infty}\bigcup_{k \geq n}^{\infty}\{x:f_k(x)>a\}$$ $$\{x: \liminf_nf_n(x)>a\}=\bigcup_{n=1}^{\infty}\bigcap_{k \geq n}^{\infty}\{x:f_k(x)>a\}$$
Also note that $$f(x)=\limsup_nf_n(x)$$ a.e so $$f$$ is measurable.