The question is from "Measure theory and probability theory (pag. 51)" by Krishna and Soumendra that lacks of a demonstration.
Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and $f:\Omega \rightarrow [0, \infty]$ a non-negative measurable function. Suppose $\{\delta_n\}$ to be a sequence of positive real numbers and $\{N_n\}$ a non-decreasing sequence of integers such that $\delta_n \rightarrow 0$, $N_n \rightarrow \infty$ and $\delta_n N_n \rightarrow \infty$ (for example $\delta_n=2^{-n}$ and $N_n=n2^n$).
Define
\begin{gather*}
f_n(\omega)=\left\{
\begin{array}{ll}
j\delta_n \ \ \text{if} \ \ j\delta_n \le f(\omega) <(j+1)\delta_n \ \ j=0,\dots,N_n-1 \\
\delta_nN_n \ \ \text{if} \ \ f(\omega) \ge \delta_nN_n
\end{array}
\right.
\end{gather*}
I need to prove that $\{f_n\}$ converges to $f$ pointwise and that each $f_n$ is measurable.
Best Answer
The preimages of $f_n$ are unions of sets of the form $\{\omega : a \le f(\omega) < b\}$ and $\{\omega : f(\omega) \ge a\}$, which are measurable sets because $f$ is measurable.
For a fixed $\omega$, we want to check that $f_n(\omega) \to f_n(\omega)$. Because $\delta_n N_n \to \infty$, we have $f(\omega) < \delta_n N_n$ for all large $n$. By the definition of $f_n$, we have $f_n(\omega) = j\delta_n$ for some $j$, so $f_n(\omega) \le f(\omega) < f_n(\omega) + \delta_n$. Taking $\delta_n \to 0$ establishes the convergence.