Here is theorem 1.17 in Rudin's Real and Complex Analysis
Let $f : X \to [0,\infty]$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that (a) $0 \le s_1 \le s_2 \le ... \le f$ and (b) $s_n(x) \to f(x)$ as $n \to \infty$ for every $x \in X$.
I am working through both Rudin's proof and this proof of the above theorem. I just want to go over a few things to make sure I understand everything.
First, I think I can see why $\varphi_n$ is a Borel function: both $[k2^{-n},(k+1)2^{-n})$ and $[n,\infty]$ are Borel sets, so the characteristic function on each is a Borel function; finally, the sum of Borel function is a Borel function, making $\varphi_n(t)$ a Borel function. Second, I think I can see that $\varphi_n(t) \to t$. Given $t \ge 0$ and $n \in \Bbb{N}$, there exists a nonnegative integer $k_n$ such that $k_n < t 2^{n} < k_n+1$ or $k_n 2^{-n} < t < k_n 2^{-n} + 2^{-n}$ and therefore $0 < t - k_n 2^{-n} < 2^{-n}$. But $\varphi_n(t) = k_n 2^{-n}$, so that taking the limit $n \to \infty$, we get $\varphi_n(t) \to t$. Doing the the same for $f(x) \ge 0$, we can show that $\varphi_n(f(x)) \to f(x)$ for every $x \in X$. Hopefully someone having greater familiarity with the subject and Rudin's book RCA will correct me if I am wrong.
However, I am at the moment having trouble seeing why $\{\varphi_n\}$ is a monotonically increasing sequence.
For an interval $I:=\left[k2^{-n},(k+1)2^{-n}\right)$, we have \begin{align*} I&=\left[2k\cdot 2^{-(n+1)},(2k+2)\cdot 2^{-(n+1)}\right)\\ &=\left[2k\cdot 2^{-(n+1)},(2k+1)\cdot 2^{-(n+1)}\right)\bigcup\left[(2k+1)\cdot 2^{-(n+1)},(2k+2)\cdot 2^{-(n+1)}\right), \end{align*} you may now see that for a typical interval of $\phi_{n+1}$, it belongs to either the left part or the right part of a typical interval of $\phi_{n}$, and if you compare the values that $\phi_{n+1}$ and $\phi_{n}$ taken over there, they are $l2^{-(n+1)}$ and $k2^{-n}$ respectively, if it were belonging to the left part, then $l=2k$, if it were right part, then $l=2k+1$, in either case, $\phi_{n+1}(x)\geq\phi_{n}(x)$ over there.