Proof explanation of Every nonnegative function can be written as the increasing limit of a sequence of simple functions

Question

Consider the following Theorem (ii)

In the part that says by passing from $f_k$ to $f_{k+1}$ each subinterval is divided in half. It follows that $f_k\le f_{k+1}$.

I can see that indeed each subinterval is divided in half but I cannot see how that implies that $f_{k}$ is monotone increasing.

Could you explain please?

In the part that states wherever $f$ is finite, we have $$0\le f-f_k\le 2^{-k}$$

Why ?

If $f$ is finite then $f_k=\frac{j-1}{2^k}$

and where does $0\le f-f_k\le 2^{-k}$ come from?

Note this is not a duplicate with If $f$ is measurable function then there exist a sequence of $φ_n$ measurable functions such that $φ_n\to f$ pointwise because the presented proofs are different.

Answer 1

Put differently, in step $k$, we partition $[0,\infty)$ as $$\tag1 [0,\infty)=[0,2^{-k})\cup [2^{-k},2\cdot 2^{-k})\cup \ldots \cup [(j-1)2^{-k},j2^{-k})\cup \ldots \cup [k-2^{-k},k)\cup [k,\infty)$$ and let $f_k(x)$ be the lower end point of the unique interval on the right of $(1)$ containing the value $f(x)$. Now note that the partition for $k+1$ is a refinement of the partition for $k$. Hence, if $f(x)\in[a,b)\subseteq [c,d)$, where $[a,b)$ is an interval of step $k+1$ and $[c,d)$ an interval of step $k$, we have $f_{k+1}(x)=a\ge c=f_k(x)$.

Regarding the second part, the claim $f(x)-f_k(x)<2^{-k}$ does not hold as generally as written, but rather only eventually, namely, when $k\ge f(x)$. For then $f(x)$ is in one of the $k2^k$ first intervals in $(1)$, and as these are of length $2^{-k}$, the claim $0\le f(x)f_k(x)<2^{-k}$ follows. Of course, the merely eventual nature of the inequality does not disturb the ultimately desired claim about convergence.

Answer 2

The proof is, in effect, doing a base-$2$ version of looking at the decimal expansions of the functions values. For example, suppose $f(x)=\pi=3.14159265\ldots$. Then we can approximate $f$ with a sequence of simple functions $g_k$ whose values never exceed $k$ and never have more than $k$ decimal digits:

$$\begin{align} g_1(x)&=1.0\\ g_2(x)&=2.00\\ g_3(x)&=3.000\\ g_4(x)&=3.1415\\ g_5(x)&=3.14159\\ g_6(x)&=3.141592\\ &\,\,\vdots \end{align}$$

Written in binary, we have $\pi=11.0010010000111111\ldots$, and so the proof's sequence increasing to the value $f(x)=\pi$ is

$$\begin{align} f_1(x)&=1.0\\ f_2(x)&=10.00\\ f_3(x)&=11.000\\ f_4(x)&=11.0010\\ f_5(x)&=11.00100\\ f_6(x)&=11.001001\\ &\,\,\vdots \end{align}$$

In general we have $f_k(x)\le f_{k+1}(x)$ because $f_{k+1}$ looks at more binary digits of the value $f(x)$, and, once $k\gt f(x)$, we have $0\le f(x)-f_k(x)\lt2^{-k}$ because $f_k(x)$ is looking at the first $k$ binary digits of $f(x)$.