Proving a non-negative measurable function is the limit of a sequence of simple functions

Question

I'm reading some notes on measure theory and am trying to understand a proof of a theorem that says:

Given a measurable function $f : \Omega \rightarrow \mathbb{R}$, there exists a sequence of isotone simple functions $(f_i)_{i \in \mathbb{N}}$ such that for each $x \in \Omega, \lim_{i\to\infty}f_i(x) = f(x)$.

The notes say they will prove this fact by proving there exists a family of sets $\{A_i\}_{i\in\mathbb{N}}$ such that $$f = \sum_{k=1}^\infty \frac{1}{k} \chi_{A_k}$$.

I understand all of the steps the proof takes after this, but I did not understand how proving such a family of sets exists proves the theorem.

Answer 1

A simple function $g$ is defined as $$g(x)=\sum_{k=1}^N a_k\chi_{A_k}$$ with $A_k$ measurable. Then your can define a sequence of simple functions $g_N$ by $$g_N(x)=\sum_{k=1}^N \frac{1}{k}\chi_{A_k},$$ i.e. basically cutting of the infinite sum. Then $$f=\sum_{k=1}^\infty \frac{1}{k}\chi_{A_k}=\lim_{N\rightarrow\infty}\sum_{k=1}^N \frac{1}{k}\chi_{A_k}=\lim_{N\rightarrow\infty} g_N$$

Answer 2

Since sums of simple functions are simple, the expression $$\sum_{k=1}^\infty \frac{1}{k} \chi_{A_k}$$ is precisely a limit of simple functions.