$\newcommand{\scrF}{\mathscr{F}} \newcommand{\scrB}{\mathfrak{B}} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}}$ Proposition: Let $(\Omega, \scrF)$ be a measurable space. A function $f: \Omega \rightarrow [0, \infty]$ is measurable if and only if it is a limit of an increasing sequence of non-negative simple functions. More precisely there exists a sequence $(s_n)_{n \geq 1}$ of simple functions such that $0 \leq s_1 \leq s_2 \dots \leq f$, and $s_n \rightarrow f$ pointwise on $\Omega$.
I'm trying to understand the proof but while I think I get the base idea of parittioning the interval $[0,n)$ into $n2^n-1$ disjoint sets and then considering preimages I'm getting bogged down in the measurability of the functions. For reference here is scratch for my attempt at a proof but this isn't the main issue yet.
Beginning of an attempt: If $s_n \rightarrow f$ pointwise on $\Omega$ then the limiting function exists and is measurable because the simple functions $s_n$ are measurable for all $n \geq 1$. To prove the converse, let $n \in N$ and define \begin{equation*} D_{k,n} = \left\{\omega \in \Omega \mid \frac{k}{2^n} \leq f(\omega) < \frac{k+1}{2^n}\right\}, \; (0 \leq k \leq n2^n-1), \end{equation*} and then because this goes up to $k = n2^n-1$ we simply set \begin{equation*} D_{n2^n,n} = \{\omega \in \Omega \mid f(\omega) \geq n\}. \end{equation*} Then the sets $D_{k,n}$ for $0 \leq k \leq n2^n-1$ correspond to the preimages under $f$ of those partitions. More precisely, note that these sets correspond to $D_{k,n} = f^{-1}([k/2^n, (k+1)/2^n))$ and $D_{n2^n,n} = f^{-1}([n, \infty])$ from which it follows that they are each measurable sets as $f$ is measurable and these are Borel sets. With this we will define a sequence of functions by $$ \phi_n = \sum_{k=0}^{n2^n-1} \left(\frac{k}{2^n}\chi_{D_{k,n}} \right)+ n\chi_{D_{n2^n, n}}. $$ Now note that if $f(\omega)<n$ it follows that $\chi_{D_{n2^n,n}}(\omega) = 0$ and moreover that there exists a unique $k'$ for which $k'/2^n \leq f(\omega) < (k'+1)/2^n$. Because of this it follows that $$ \phi_n(\omega) = \sum_{k=0}^{n2^n-1} \frac{k}{2^n} \chi_{D_{k,n}} = \frac{k'}{2^n} $$ since $\chi_{D_{n,k}} = 0$ for all $k \neq k'$. Because $\phi_n(\omega) = k'/2^n$ and $f(\omega)$ lives within the interval $[k'/2^n, (k'+1)/2^n)$ which has length $1/2^n$ it follows that when $f(\omega) < n$ we have $$ |f(\omega) - \phi_n(\omega)| < \frac{1}{2^n}. $$ On the other hand if $f(\omega) > n$ it follows that $D_{k,n} = 0$ for every $0 \leq k \leq n2^n-1$ and $D_{n2^n,n} = 1$ so that $\phi_n(\omega) = n$...
My main issue here is the following. If $(\Omega, \scrF)$ is a measurable space and $f: \Omega \rightarrow [0, \infty]$ what is the $\sigma$-algebra on the codomain? On wikipedia they write that it's the restriction of $\scrB(\R^*)$ to $[0,\infty]$ but what does that mean exactly? The main issue in my mind is that for $D_{k,n}$ to be a measurable set I believe I would need $[\frac{k}{2^n}, \frac{k+1}{2^n})$ to be Borel. But in my mind to justify that a set of the form $[a,b)$ is Borel I could write that for $a<b$ $$ [a,b) = [-\infty, a)^c \cap [-\infty, b) = [a,\infty] \cap [-\infty, b), $$ and since $[-\infty, a)$ and $[-\infty, b)$ are open and so $[a,b)$ is Borel. But I needed to have access to intervals of the form $[-\infty, a)$ which I won't have if I'm just considering $[0,\infty]$ so what do they mean when they say $\scrB(\R^*)$ restricted to $\R^+$? Do they mean the minimal sigma algebra generated by $[0,\infty]$? Thanks in advance for your help.
Your proof is correct, and actually any interval $[a, b)\subseteq [0,+\infty]$ is a Borel set, so that each $D_{k,n}=f^{-1}([\frac{k}{2^n}, \frac{k+1}{2^n}))\subseteq \Omega$ is measurable because $f$ is a measurable function.
More precisely, the idea of restricting the Borel $\sigma$-algebra to $[0,+\infty]$ is that we want to keep the same $\sigma$-algebra we began with, but of course ignoring Borel sets that are not entirely within the subset $[0,+\infty]$ we are interested in.
In general, we have the simple following fact: if $\mathcal{A}$ is a $\sigma$-algebra on $\Omega$ and $S\subseteq\Omega$ is a subset, then $$\mathcal{A}_S=\{A\cap S | A\in \mathcal{A}\}$$ is a $\sigma$-algebra on $S$. Moreover, if $S$ is itself in $\mathcal{A}$, then $\mathcal{A}_S=\{A\in\mathcal{A}|A\subseteq S\}$.
Especially in this second case, we can sensibly call $\mathcal{A}_S$ the restriction of $\mathcal{A}$ to $S$, since it contains exactly all the sets in $\mathcal{A}$ that are contained in $S$.
In general, it is easily seen that $\mathcal{A}_S$ is the coarsest $\sigma$-algebra on $S$ which makes the inclusion map $i:S\hookrightarrow \Omega$ measurable. If you are familiar with topology, you may recognize the analogy with the definition of subspace topology.
Now, you know that there is a $\sigma$-algebra on $\overline{\mathbb{R}}$ called the Borel $\sigma$-algebra, and it contains any interval of the form $[a,b)$. If you restrict yourself to the subset $[0,+\infty]$ with the definition given above, any $[a,b)\subseteq [0,+\infty]$ is still a Borel set in the restricted σ-algebra.
The nice property of this definition is that, for every measurable space $(X, \mathcal{M})$ and every function $f:X\to S$, we can see $f$ as a function which takes values in $\Omega$, since $S$ is a subset of it. More precisely we should consider $\overline{f}=i\circ f:X\to \Omega$. Now the property is (universal property of restricted $\sigma$-algebra): if we put on $S$ the restricted σ-algebra $\mathcal{A}_S$, then $$f \text{ is measurable } \iff \overline{f} \text{ is measurable}$$
Therefore, if you do not like this fact of restricting the $\sigma$-algebra, you could equally consider your $f$ (as suggested in the comments) as a function on the whole $\overline{\mathbb{R}}$ (technically, you are considering $\overline{f}=i\circ f$, where $i$ is the inclusion $[0,+\infty]\hookrightarrow \overline{\mathbb{R}}$), and you know by the universal property that $\overline{f}$ is measurable (since $f$ is measurable), and therefore $f^{-1}([a,b))=\overline{f}^{-1}([a,b))$ is measurable in $\Omega$ as $[a,b)$ is a Borel set in $\overline{\mathbb{R}}$.