# Proving a set is $\mathcal{F}$-measurable

I am studying for a math exam, and I am unsure about how to do this problem I have for practice.

Let $$f_n$$, $$n\geq 1$$ and $$f$$ be measurable functions on a measurable space $$(\Omega, \mathcal{F})$$. Show that the set $$\{\omega : \lim_{n\to\infty} f_{n}(\omega) = f(\omega)\}$$ is $$\mathcal{F}$$-measurable.

I know that I need to show the inverse is in the set as well. But I am not so sure how to. I've learned about about quite a few theorems and maybe I am overcomplicating it, but I don't know how to do this problem.

I will really appreciate any help to help me prepare.

## 2 Answers

What do you mean by "inverse"? You need to show that the set is in $$\mathcal{F}$$, this is the definition of a measurable set. Well, that's easy.

$$\{\lim_{n\to\infty} f_n=f\}=\{\omega: \forall\ (\epsilon\in\mathbb{Q}\cap (0,\infty))\ \exists (n_0\in\mathbb{N})\ \forall (n\geq n_0) [|f_n(\omega)-f(\omega)|<\epsilon]\}=$$

$$=\cap_{\epsilon\in\mathbb{Q}\cap (0,\infty)}\cup_{n_0=1}^\infty\cap_{n=n_0}^\infty\{|f_n-f|\in (-\epsilon,\epsilon)\}$$

So this is a set which we obtain from countable intersections and unions of measurable sets, hence it is measurable. Of course we use the fact that $$f_n$$ and $$f$$ are all measurable functions, which implies that for all $$n\in\mathbb{N}$$ the function $$|f_n-f|$$ is measurable as well.

$$f(x)=\limsup_nf_n(x)=\liminf_nf_n(x)$$

Prove that $$\limsup_nf_n$$ or $$\liminf_nf_n$$ is measurable.