For each $n$ in $\mathbb{N}^*$, let $$d_n(r) = \sum_{k=1}^{n2^n} \frac{k-1}{2^n} 1_{[\frac{k-1}{2^n}, \frac{k}{2^n})} (r)+ n 1_{[n, \infty]}(r), r\in \mathbb{\bar{R}}_+,$$
Then, each $d_n$ is an increasing right-continuous simple function on $\mathbb{\bar{R}}_+$ and $d_n(r)$ increases to $r$ for each $r$ in $\mathbb{\bar{R}}_+$ as $n \rightarrow \infty$.
This is given as an lemma in my textbook without proof. It doesn't look obvious to me at all. Even understanding the function itself is hard for me.
Can someone kindly explain to me the function, how to interpret and understand it and prove the lemma?
It seems to be that $1_{[\frac{k-1}{2^n}, \frac{k}{2^n})}(r)$ is satisfied only when $k = n2^n$ which is the last case of the summation. Why do we need the summation then, instead of taking the last term only?
In terms of the greatest integer $[x]$ not exceeding $x$ you can write $d_n (r)=\frac {[2^{n}r]} {2^{n}}$ if $r \leq n$ and $d_n(r)=n$ if $r>n$. Proof of the fact that $\frac {[2^{n} x]} {2^{n}} $ increases to $x$ as $ n$ increases to $\infty $ for every $x\geq 0$: we have $x-\frac 1 {2^{n}} =\frac {2^{n} x-1} {2^{n}} <\frac {[2^{n} x]} {2^{n}} \leq \frac {2^{n} x} {2^{n}}=x $. By the Squeeze Theorem $\frac {[2^{n} x]} {2^{n}} \to x$. Next we prove that $2[y] \leq [2y]$ for all $y \geq 0$: if $n \leq y <n+1$ then $2n \leq 2y <2n+2$ so either $2n \leq 2y <2n+1$ or $2n+ \leq 2y <2n+2$. Hence $[y]=n$ and $[2y] =2n$ or $[2y]=2n+1$. It follows that $[2y] \geq 2n =2[y]$ in both cases. Taking $y=2^{n}x$ we get $\frac {[2^{n} x]} {2^{n}} \leq \frac {[2^{n+1} x]} {2^{n+1}} $. This completes the proof.