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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.

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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.

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

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

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