Let $(X,\mathscr{A},\mu)$ be a measure space. The Riesz Theorem in real analysis shows that, for $A\in\mathscr{A}$, functions $f,f_{n}:A\to\mathbb{R}$ are $\mu$-measurable, if sequence $\{f_n\}$ is $\mu$-convergent to $f$, then there exists a subsequence $\displaystyle{\{f_{n_{k}}\}}$ of $\{f_n\}$ such that $f_{n_{k}}$ convergent to $f$ $\mu$-almost everywhere.
I know the condition "$\mu$-almost everywhere convergence" is necessary, but I wonder if there is a counterexample without this condition. That is to say, are there such functions $f$ and $f_n$ that satify $f_n$ is $\mu$-convergent to $f$, but do not exist any subsequences $\{f_{n_{k}}\}$ pointwisely convergent to $f$ in $A$.
If I understand the question correctly: "$\mu$-convergence" is what's known perhaps more commonly as convergence in measure; you remind us that $\mu$-convergence implies convergence of a subsequence almost everywhere, and wonder whether there's an example showing that it does not imply that some subsequence converges at every point.
The simplest answer is that there must be such an example, since $\mu$-convergence is not affected by sets of measure zero. For example, say $\mu$ is Lebesgue measure. Say $(a_n)$ is a sequence of scalars with no convergent subsequence, and let $$f_n(x)=\begin{cases}0,&(x\ne0),\\a_n,&(x=0).\end{cases}$$Then $f_n$ certainly tends to zero in measure, but no subsequence converges at the origin.