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I have a question about the set $E_{kn}$, what is this? What are the parameters $n,k$ for? I guess actually $k$ is the real parameter because $n$ is fixed natural number. How are we cutting $f(x)$ here?

This was taken from Bartle

Lemon
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    $n,k$ are defined right after the word "Proof". – coffeemath Aug 01 '17 at 04:00
  • @coffeemath, yeah but what does the set $E_{kn}$ even look like? What is this strange object? – Lemon Aug 01 '17 at 04:01
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    We're cutting the nonnegative half of the real line into $n\cdot2^n$ intervals of length $2^{-n}$ each (so they cover the interval from $0$ up to $n$) plus the half-line from $n$ upward. $E_{kn}$ is the set of those points $x\in X$ that $f$ maps into interval number $k$ (where the short intervals are numbered from $0$ through $n2^n-1$ and the half-line at the end counts as interval number $n2^n$), – Andreas Blass Aug 01 '17 at 04:03

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That's fairly awful notation: a plain $X$ and a bold $\mathbf X$ that are pretty difficult to distinguish. Also $E_{kn}$, ugh! $E_{k,n}$ would be much better. Of course $k$, $n$ and $E_{k,n}$ are defined clearly enough within the proof.

But the $E_{k,n}$ are just a means to the end of defining the $\newcommand{\vp}{\varphi}\vp_n$. Essentially $\vp_n$ is a "rounding-down" of $f$ in some manner. It is given that $f$ is positive. When $f(x)>n$ round it down to $n$. If $0\le f(x)\le n$ round it down to the nearest multiple of $2^{-n}$. This gives $\vp_n$. The point is that $\vp_n$ is a simple function with level sets in $\mathbf X$ (whatever that is) and the $\vp_n$ increase to the limit $f$.

Angina Seng
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Basically the image of $f$ is cut into the sets $\cup_k^{n2^n -1} [k/2^n, (k-1)/2^n] \cup [n, \infty)$ and the image of $f$ under $[k/2^n, (k-1)/2^n]$ is partitioned into $1/2^n$ pieces. So the length of $[k/2^n, (k-1)/2^n]$ is $1/2^n$. So the simple functions $\phi$ are

$\phi_n(x) = n$ if $f(x) \in [n,\infty)$ else $\phi_n(x) = k/2^n$ (because $[k/2^n, (k-1)/2^n] \cap [n, \infty) = \emptyset$

Lemon
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