Let $X$ be a measurable space. A measure $\mu$ on $X$ is called $\sigma$-finite if there exist measurable subsets $A_{1},A_{2},\ldots\subset X$ such that $\bigcup_{j=1}^{\infty}A_{j}=X$ and $\mu(A_{j})<\infty$ for all $j=1,2,\ldots$. A function $f\colon X\to\mathbb{C}$ is called strongly $\mu$-measurable if there exists a sequence $(f_{n})$ of $\mu$-simple functions (i.e. $f_{n}$ is a finite linear combination of indicator functions $1_{A}$ with $\mu(A)<\infty$), such that $f_{n}(x)\to f(x)$ for almost every $x\in X$.
I want to prove that the following statements are equivalent:
(i): $\mu$ is $\sigma$-finite.
(ii): $1_{X}$ is strongly $\mu$-measurable.
The implication (i) $\implies$ (ii) is easy: just put $f_{n}:=1_{\bigcup_{j=1}^{n}A_{j}}$. But how do I prove the converse? Any help would be appreciated!
First off, a remark: if you have $(A_n)_n$ of finite measure such that $\mu(X \setminus \bigcup A_n) = 0$, then interchanging $A_1$ with $A_1 \sqcup X \setminus \bigcup_{n \geq 1} A_n$ we have that $X$ is $\sigma$-finite.
Take $f_n \to 1_X$ pointwise a.e. with each $f_n$ simple and put $A_n = \{f_n=1\}$. (I adopted the notation from the comments, which I found better than the ad-hoc one I had used).
Observe that $x \in A_n$ if and only if $f_n(x) = 1_{A_n}(x) = 1$, and $f_n(x) \to 1_X(x) \equiv 1$ if and only if $(f_n(x))_n$ is eventually constant, i.e. if $x \in \bigcup_m \bigcap_{n > m} A_n = \lim \inf A_n$.
Since $f_n \xrightarrow{(p)} 1_X$ a.e., we see that $\mu(X \setminus \lim \inf A_n) = 0$. Writing $B_m = \cap_{m > n} A_n$, we have that $\mu(B_n) \leq \mu(A_n) < \infty$ and $\mu(X \setminus \bigcup B_m) = 0$, proving the claim.