why is $f_{n}\rightarrow f \; \text{point-wise} \iff \limsup_{n\rightarrow\infty}f_{n}=\liminf_{n\rightarrow\infty}f_{n}=f$

Question

in a proof of a lemma that states the following : $f_{n}\rightarrow f \; \text{point-wise} \implies f \; \text{is measurable} $

the author proceeded in 3 steps :

1. proving that $\sup_nf_n\;\text{&}\; \inf_nf_n$ are both measurable

2. proving that $\lim\sup_nf_n\;\text{&}\; \lim\inf_nf_n$ are both measurable

3. concluding from the following claim that $ f \; \text{is measurable}$

   $$\limsup_{n\rightarrow\infty}f_{n}=\liminf_{n\rightarrow\infty}f_{n}=f$$

I know that $\lim f_n = f$ but couldn't prove the above statement that I've never come across till this day.

if anybody could prove or explain why does that equivalence hold it'd be great.

thanks!

Answer 1

If we prove that $\inf_n f_n(x)$ and $\sup_nf_n(x)$ are measurable then we can can prove that $\lim\sup_n f_n(x)=\inf_n \sup_{m \geq n}f_m(x)$ and $\lim \inf_n f_n(x)=\sup_n \inf_{m \geq n}f_m(x)$ are measurable.

Thus the function $f(x)=\lim\sup_n f_n(x)=\lim\inf_n f_n(x)=\lim f_n(x)$ will be measurable,because the set $$\{x:\limsup_n f_n(x)>a\}$$ $$=\{x:\lim \inf_n f_n(x)>a\}$$ $$=\{x:\lim_nf_n(x)\}$$ $$=\{x:f(x)>a\}$$ would be measurable $\forall a \in \Bbb{R}$