Let $f_n$ be a sequence of measurable functions on $E$ that converges to the real-valued $f$ pointwise on $E$. Show that $E = \bigcup_{k=1}^\infty E_k$, where for each index $k$, $E_k$ is measurable, and $f_n$ converges uniformly to $f$ on each $E_k$ if $k > 1$ and $m(E_1)=0$.
This problem was asked on MSE already (see this). But I am wondering whether the following proof, which seems slightly different from the one already linked, establishes the theorem?
Let $\epsilon > 0$. Define $E_1 = E \cap \Bbb{Q}$, and therefore $m(E_1)=0$, and define $E_{k} := \{x \in E \mid |f(x)-f_n(x)| < \epsilon, ~ \forall n \ge k-1 \}$ for $k > 1$. Clearly $E_1$ is measurable, since it has zero measure, and note that $E_k = \bigcap_{n =k-1}^\infty \{x \in E \mid |f(x)-f_n(x)| < \epsilon \}$ is measurable since it is a countable intersection of measurable sets. Let $x \in E$. Then by pointiwise convergence, there exists an $N \in \Bbb{N}$ such that $|f(x)-f_k(x)| < \epsilon$ for every $k \ge N$. If $N=1$, choose $k > 1$, and therefore $x \in E_k$; else $x \in E_N$ works. Hence $E = \bigcup_{k=1}^\infty E_k$.
Let $k > 1$, and now we prove uniform convergence on $E_k$. Take $k$ to be our "index of convergence." Let $n \ge k$ and $x \in E_k$. Then $x \in E_n$ for every $n \ge k$ which implies $|f(x)-f_n(x)| < \epsilon$ for every $n \ge k$. This proves that $f_n \to f$ uniformly on $E_k$.
So, how does that sound? This is sort of a strange problem. I feel like I am cheating by taking $E_1 = E \cap \Bbb{Q}$.
EDIT:
I think there may be one mistake in the second part of the proof: I think we need to take $k-1$ as the "index of convergence" instead of $k$. Not sure at the moment, though.