# Edited: Defining a measurable pointwise limit for a sequence of measurable functions.

Edit: I edited the post considerably to focus attention to the primary question

Let $\left(X,\mathcal{F},\mu\right)$ be a measure space and let $f_{n}:\left(X,\mathcal{F}\right)\to\mathbb{R}$ be a sequence of $\mathcal{F}$ -measurable functions. Suppose $\mathbb{R}$ is equipped with the Borel $\sigma$ -algebra. I know that $$\left\{ x\in X\,|\,\lim\limits _{n\to\infty}f_{n}\left(x\right)\in\mathbb{R}\right\}$$ is a measurable set. Define the following function: $$f\left(x\right):=\begin{cases} \lim\limits _{n\to\infty}f_{n}\left(x\right) & \mbox{if the limit exists and is finite}\\ 0 & \mbox{otherwise} \end{cases}$$ I want to show this is an $\mathcal{F}$-measurable function but I'm not sure how. If the sequence converged pointwise everywhere then $f\left(x\right)={\displaystyle \limsup_{n\to\infty}f_{n}\left(x\right)}$ and that suffices. However, if the convergence is not everywhere I'm not sure what to do. Help would be appreciated!

It is actually true for all measures that $\limsup f_n$ is measurable:

To see this, we first show that $h=\sup f_n$ is measurable.

Let $a \in \mathbb{R}$, and let $E_a=\{ x \in E: h(x) \leq a \}$ and we show it is measurable.

Then $E_a=\cap_n \{ x \in E: f_n(x) \leq a \}$, so it is measurable.

Now we can represent $\limsup f_n$ as a $\sup$ of measurable functions, and we are done:

Define $g_n=\sup_{k \geq n} f_k(x)$, so these are measurable functions, and $\limsup f_n=\inf g_n=-\sup(-g_n)$. $\blacksquare$

• This answer was made irrelevant by my editing. – Serpahimz Feb 3 '14 at 23:40