How the function $f_n$ is a measurable function?

Question

My professor has defined a function and told us that the function is measurable. I cannot find any suitable explanation how this function is measurable? It is defined as,

Let, $E$ be a measurable set and $f:E\rightarrow [0,\infty]$ be a measurable function. For each $n\in \mathbb{N}$ we define $f_n:E\rightarrow [0,\infty)$ by,

$f_n= \begin{cases} 0,& if & 0\leq f(x)<\frac{1}{2^n} \\ \frac{1}{2^n}, & if & \frac{1}{2^n}\leq f(x)<\frac{2}{2^n} \\ \frac{2}{2^n}, & if & \frac{2}{2^n}\leq f(x) <\frac{3}{2^n} \\ \dots & \dots & \dots \\ \frac{n2^n-1}{2^n}, & if & \frac{n2^n-1}{2^n}\leq f(x) <n \\ n, & if & f(x)\geq n \end{cases}$

Moreover if possible please show that the sequence $\{f_n\}_{n\in \mathbb{N}}$ is monotonic increasing!

The function is too much complicated. Any help will be appreciated. Thanks in advance.

Answer 1

For the measurability part, as pointed in the comments section, we can just use the definitions or, alternatively, consider the following argumennt:

Let $\chi_k:E\to\{0,1\}$, $k=0,1,\dots,n2^n-1,n2^n$ be the characteristic functions of the measurable sets $I_k$, where: $$\begin{align} I_k&=f^{-1}\left(\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)\right),\ \text{for }k=0,1,\dots,n2^n-1\\ I_{n2^n}&=f^{-1}\left([n,+\infty)\right). \end{align}$$ Then, since $I_k$ are pairwise disjoint, we have that: $$f_n=\sum_{k=0}^{n2^n}\frac{k}{2^n}\chi_k$$ and, hence, $f_n$ is measurable as a finite linear combination of measurable functions.

For the monotonicity part, consider the following hint:

I tried to play with line thickness and the colours. Can you make now this intuitive idea strict mathematic arguments?>