Egoroff's theorem for non $\sigma$-finite measure spaces, but each $f_n$ is integrable

Question

This question is from a past qualifying exam, and therefore I am interested in tips/advice/approach over the actual proof.

Problem.

Fix a measure space $(X,\mathcal{A},\mu)$. Fix $\{f_n\}_{n\in\mathbb{N}}\subseteq L^1(X,\mathcal{A},\mu)$. Assume $f_n$ converge pointwise $\mu$-a.e. to a measurable function $f$. Show that there exist measurable sets $H,E_k$ ($k\in\mathbb{N}$) such that \begin{align} &X=H\cup\bigcup_{k\in\mathbb{N}}E_k,\\ &\mu(H)=0,\\ &f_n\to f\text{ uniformly on each }E_k. \end{align}

My instincts tell me to first let $H$ collect points so that $f_n\to f$ pointwise on $X\setminus H$. Then show $\{x:f(x)\ne0\}$ is sigma finite (to apply regular Egoroff theorem in this context). But then that leaves the possibly non sigma finite set $\{x:f(x)=0\}=:Z$ which is where I am stuck.

(1) How does one work with the (possibly) non sigma finite set $Z$ to get uniform convergence?

Answer 1

There is one set $A$ of sigma finite measure such that $f_n=0$ for all $n$ and $f_n \to f$ outside $A$. Hence $f$ also vanishes outside $A$. So non-sigma finiteness is no problem at all.