I'm trying to prove that given a convergent in measure sequence of measurable function $\{f_n\}_n$ from a measure space $X$ to $\mathbb{C}$, if $f$ is (any of) its limit in measure, then $f$ is also measurable.
$\{f_n\}_n$ converges in measure to $f$ means that $$\lim_n\mu(\{x\in X : \quad |f_n(x) - f(x)|>\epsilon\})=0$$
To me the main dificulty is that with this type of convergence I have no idea of how $f$ looks like at any point.
I'm not sure if this is supposed to be proved using the definition of measurable function or if there is some "smart trick" to use. In the first case I take an open set $A\subset\mathbb{C}$ and I want to conclude $f^{-1}(A)$ is measurable. I think that maybe $f^{-1}(A)$ can be expressed in terms of the functions of the sequence but I don't see how.
Any hint or attept is welcome, I've asked this question 8 months ago when I was taking the course of measure theory but I didn't recieve any atention, now I'm preparing for the exam and I can't still solve it.
Because the sequence converges in measure, there is a subsequence that converges pointwise, $\mu$-a.e. That is, there is a subsequence $n(k)$ and a measurable set $N$ with $\mu(N)=0$ such that $\lim_k f_{n(k)}(x) =f(x)$ for all $x\in X\setminus N$. Define $g(x):=\limsup_k \mathfrak{R}(f_{n(k)}(x))$ and $h(x):=\limsup_k\mathfrak {I}(f_{n(k)}(x))$ ($\mathfrak{R}$ and $\mathfrak{I}$ denote real and imaginary parts). Then $L:=g+ih$ is a complex-valued measurable function such that $f=L$, $\mu$-a.e. This is the best you can say (in general) about the measurability of $f$; if $\mu$ is complete, then $f$ is measurable.