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$(\Omega,\Sigma,\mu)$ is a probability measure space, and $X$ is a Banach space.

$f_n:\Omega\rightarrow X$ are simple functions for each $n\in\mathbb{N}$, and$f:\Omega\rightarrow X$ satisfies $f_n\rightarrow f$ almost everywhere (so we have $f$ is $\mu$-measurable in some sense.)

Can I choose a sequence of countably valued functions $(g_n)$ such that $\Vert g_n-f\Vert \rightarrow 0$ uniformly almost everywhere? i.e., for each $k>0$, $\exists N>0$ such that $\Vert g_n-f\Vert<\frac{1}{k} \forall n>N$ a.e.

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    I don't see the relevance of $(f_n)$ here – 2017-01-20
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    I think it is possible to construct $g_n$ from $f_n$ but I am stuck – 2017-01-20
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    Oh, I see. f is not simple, sorry! – 2017-01-20

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Let $A_{n,i}=[i/n,(i+1)/n )$, this is a covering of $\Bbb R$ via countable intervals of length $1/n$. Define $$g_n(x)=\sum_{i\in\mathbb Z} \frac in\ \chi_{f^{-1}(A_{n,i})}(x) $$ ie $g_n(x)=\frac1n\lfloor nf(x)\rfloor$. $g$ is countable valued and $$|f(x)-g_n(x)|≤\frac1n$$ so $g_n$ converges uniformly to $f$.

The measure does not enter, so unless you want some other kinds of restrictions on $g_n$ you do not care about the properties of $f$ wrt $\mu$.

However in the case that $X$ is a Banach space it is needed. Let $S=\bigcup_n \mathrm{im}(f_n)$. Since the $f_n$ are simple functions $\mathrm{im}(f_n)$ is finite for all $n$ and $S$ is countable.

Since $f_n\to f$ pointwise almost everywhere there exists a measure zero set $N\subset \Omega$ so that for all $x\in \Omega-N$ and $n\in\Bbb N$ you have a $y\in S$ so that $\|y-f(x)\|≤\frac1n$.

With this information define $g_n(x)$ to be such a $y$ for $x\notin S$ and $0$ for $x\in S$. $g_n$ is countably valued (taking values in $S\cup \{0\}$) and $\sup_{x\in\Omega-N}\|g_n(x)-f(x)\|≤\frac1n$.

So you have uniform convergence "almost everywhere", where the almost everywhere is to be understood in the sense of the last equation.

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    You mean the result I asked is not relevant to $f_n$, but still correct. I think your answer is only for the case $X=\mathbb{R}$, can I just replace $A_{n,i}$ by $\{x\in X:\frac{i}{n}\le\Vert x \Vert<\frac{i+1}{n} \}$ ? – 2017-01-23
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    @CSH I did not see that we are considering functions into an arbitary banach space. This way I think we will need that $f$ is an ae limit of simple functions. – 2017-01-23
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    OK, thanks a lot! – 2017-01-23
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    @CSH I wrote up the proof in the banach space case in the answer. It uses countable choice, but we live in the future so nobody is bothered by that anymore. I'm sure there is a more constructive way to do it too. – 2017-01-23