This is part of a problem of showing that there exist a sequence of simple functions, $s_n$ that converges to a measurable function $g$.
For a fixed integer $n$, define $ s_n(x) = \begin{cases} \frac{k}{2^n}, & \mbox{if } \frac{k}{2^n}\leq g(x) \lt \frac{k+1}{2^n}~,k=0,1,2,\ldots n2^n-1 \\ n, & \mbox{if } g(x)\geq n \end{cases} $
I want to show that $s_1\leq s_2\leq s_3\leq \ldots $
Attempt:
On the interval $g(x) \geq n$, $s_{n+1} \geq n$. So on this interval, $s_{n+1} \geq s_n$.
If $g(x) < n$ , then $\frac{k}{2^n}\leq g(x) \lt \frac{k+1}{2^n},k=0,1,2,\ldots n2^n-1 $. I need some help with this part.
Thanks.