I know that this question was asked before but I have a doubt in a technique that I usually use, I will say what it is after.
The Problem: Prove that for every Lebesgue measurable function $f$ there exists a measurable Borel function $g$ such that $f=g$ almost everywhere.
All functions that we consider have $\mathbb{R}^n$ as domain and extended real numbers as codomain or a subset of it. Let $m$ be the Lebesgue measure.
I already prove that for every simple function (here simple functions are measurable) the problem is true.
Let $f$ be a measurable nonnegative function. There exists a sequence $ \{ \varphi_k \}_{k=1}^\infty$ of nonnegative simple functions such that \begin{align*} \varphi_1 \leqslant \varphi_2 \leqslant \varphi_3 \leqslant \cdots \end{align*} and \begin{align*} f(x) = \lim_{k \rightarrow \infty} \varphi_k(x) \end{align*}
For what I know from before, for every $\varphi_k$ there exists a Borel measurable function $g_k$ such that $\varphi_k = g_k$ almost everywhere. Let's do \begin{align*} Z_k = \{ \varphi_k \neq g_k \} \end{align*} then $m(Z_k) = 0$.
Now consider the next function: \begin{align*} g(x) = \lim_{k \rightarrow \infty} \inf g_k(x) \end{align*} and $g$ is a Borel measurable function. We want to proof that $f = g$ a.e. For that \begin{align*} x \in \bigcap_{k=1}^\infty Z_k^c & \Rightarrow \forall k \in \mathbb{N}, \ \varphi_k(x) = g_k(x) \\ & \Rightarrow \lim_{k\rightarrow \infty} \varphi_k(x) = \lim_{k \rightarrow \infty} \varphi_k(x) = \lim_{k \rightarrow \infty} \inf g_k(x) \\ & \Rightarrow f(x) = g(x) \end{align*}
And my doubt is if my conclusion is correct: $f(x) \neq g(x) \Rightarrow x \in \bigcup_{k=1}^\infty Z_k$. In words, $f$ and $g$ are different when $x$ is in a subset of $\bigcup_{k=1}^\infty Z_k$, wich has zero measure because the Lebesgue measure is complete (and then are equal a.e).
I know that logically speaken my conclusion is true, but I don't know, is like I "feel" something is wrong.