Rudin Theorem 1.17 1.17 THEOREM
Let $f: X \to [0, \infty)$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that
(a) $0 \le s_1 \le s_2 \le \dots \le f.$
(b) $s_n(x) \to f(x)$ as $n \to \infty \forall x$
PROOF Put $\delta_n = 2^{-n}$. To each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k \delta_n \le t < (k + 1) \delta_n$. Define
$
\phi_n(t)=
\begin{cases}
k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n \\
n \space \space \text{if} \space n \le t < \infty \\
\end{cases}
$
Each $\phi_n$ is then a Borel function on $[0,\infty)$, 
$t~\delta_n< \phi_n(t)\le t \space \space \space \space \text{if} \space \space 0 \le t <n$
$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ and $\phi_n(t) \to t$ as $n \to \infty$ It follows that the functions
$s_n = \phi_n (f)$
satisfy (a) and (b); they are measurable, by Theorem 1.12(d). $\square$
$\textbf{Question}$: Why can't we set $\phi_n(t)= n$ ? What would we lose? 
$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le  t$ is only true for $t<n$. I just don't see what we gain by having $k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n $
 A: What would we lose? Well, we need to keep our eyes on the prize: approximating $f$ by simple functions. The $k_n(t)\delta_n$ part is what is actually doing it. 
Imagine filling up a glass with some ice in the bottom, ripe for filling. As you pour water in, ignoring displacement there will be a level it would go up to, but in some spots it will be "stopped" by the ice. 
That is whats going on here essentially, you are trying to fill up to level "n" but the function gets in the way (the $k_n(t)\delta_n$ part). But, as you fill up more and more, the degree to which you are approximating $f$ gets better. 
A good picture of what is going on can be found in Folland, chapter 2. 
A: The construction partitions a subset of the range of $f$, i.e. the values of $f$ that fall in the interval $[0,n]$.  The definition of $\phi_{n}$ ensures that the interval goes to $[0,\infty)$ (i.e. covers the entire range of $f$) while the requirement $\phi$ be equal to $k_{n}(t)\delta_{n}$ ensures that the range is partitioned finer and finer (in particular, at a "faster rate" than the interval is being extended).  What you lose by just setting $\phi$ equal to $n$ is the approximation of $f$!
