Proof explanation of Every nonnegative function can be written as the increasing limit of a sequence of simple functions 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.
 A: 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 $f(x)$.
A: 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.
