Let $f: X \rightarrow [0,\infty]$ be a measurable function on a measure space. Show that there exists a squence $\{A_{j} \}$ of measurable sets such that $$f(x) = \sum_{j=1}^{\infty} \frac{1}{j}\chi_{A_{j}}(x).$$
My idea is to construct $A_{j}$ as a union of pre-images of intervals of the form $[a_{j}, \infty)$, where $a_{j}$ are some real numbers. What I get does not come out compactly. I will appreciate a hint or a solution (this is a part of qual prep)
Here is a hint.
Note that $\varphi_n(x) = \sum_{j=1}^n \frac{1}{j} \chi_{A_j}(x)$ is non-decreasing in $n$. Since you want $\lim_{n \to \infty} \varphi_n(x) = f(x)$, necessarily, $f(x) - \varphi_n(x) \geq 0$ for all $x$ and $n$. Further, $\varphi_{n+1}(x) = \varphi_n(x)$ if $x \notin A_{n+1}$ and $\varphi_{n+1}(x) = \varphi_n(x) + \frac{1}{n+1}$ if $x \in A_{n+1}$.
If we assume that $f(x) - \varphi_n(x) \geq 0$ for all $x$, for what set $A_{n+1}$, would we then also have $f(x) - \varphi_{n+1}(x) \geq 0$, but $A_{n+1}$ is as large as possible?