How can I prove like this

Question

Let $(X,S)$ be a measurable space, and $f : X \to[0, +\infty]$ be a nonnegative function ($f \ge 0$). Then the function $f$ is measurable if and only is if there is an increasing sequence of simple positive function $ψ_n$ Converges to $f$

Answer 1

Suppose that $f$ is measurable. For each $n\in \mathbb{N}$ divide the interval $[0,2^n]$ into $2^{2n}$ subintervals of length $2^{-n}$, say \begin{eqnarray} I_{k,n} &=& (k2^{-n},(k+1)2^{-n}], k = 0,1,\dots , 2^{2n}-1\\ J_n &=& (2^n,+\infty]. \end{eqnarray} We can now define a step function as follows:

$f_n = \sum_{k=0}^{2^{2n}-1}k2^{-n}\chi_{f^{-1}(I_{k,n})} + 2^n \chi_{J_n}$.

This sequence of step (simple) functions will converge to $f$ pointwise if $f$ is unbounded and uniformly if $f$ is bounded.

For the converse, show that if $f_n$ is a sequence of bounded measurable positive functions, then $\sup_n f_n$ and $\limsup_n f_n$ are measurable functions as well.