Let's split up the segment $[0, \infty]$ into chunks. For $k=1: [0, 1)(1, \infty]$. For $k = 2: [0,.5)[.5,1)[1, 1.5)[1.5, 2)(2, \infty]$ and so on .. thus we have an ever increasing resolution and ever increasing coverage of the nonnegative extended real line.
Now define the chunk $A_{k,r} = f^{-1}([k - 1 + \frac{r-1}{n}, k-1 + \frac{r}{n}))$. Geometrically it's the $r$th subblock inside of the $k$th major block, and $n$ is the index, discussed below.
Now define the simple function $s_n = \sum_{k,r = 1}^n \inf_{x \in A_{k,r}}f(x) \chi_{A_{k,r}}$.
Now I have $(a)$ of Rudin Theorem 1.17. (b) is that $s_n(x) \to f(x)$ as $n \to \infty$, for every $x \in X$.
I have that if $|f(x) - s_n(x)| \geq \epsilon$ then for some $(k,r)$ such that $A_{k,r} \ni x$ we have $|f(x) - \inf_{x \in A_{k,r}} f(x)| \geq \epsilon$.