This is a very broad question about two conceptions of a measurable function. Let $(X, \mathcal X)$ be a measurable space, and let $f$ be a real-valued function of $X$. Throughout the post, I assume that $\mathbb R$ is equipped with its Borel $\sigma$-field.
According to the standard conception of measurability, which I'll call the measure-theoretic conception, $f$ is measurable just in case $\{x: f(x) \leq r\} \in \mathcal X$ for every $r \in \mathbb R$. (There are several well known equivalents of this definition of course. The one that carries over to arbitrary measurable spaces is: $\{x: f(x) \in B\} \in \mathcal X$ for every Borel subset $B$ of $\mathbb R$.)
Why is this a natural definition? In measure theory we want to "ask questions" about $f$ and "answer" them using a measure $\mu$ on $(X, \mathcal X)$. For instance, we might want to ask, How big is the set of points on which $f$ vanishes? In other words, we want to evaluate $$\mu\big(\{x: f(x) = 0\}\big).$$ And in order to do that, we of course need the set $\{x: f(x)=0\}$ to be in the domain of $\mu$. The measure-theoretic conception ensures that this is the case.
There's another conception of measurability that isn't motivated by thinking about measures. I'll call this conception information-theoretic (perhaps there's a better name for this; this is just the first thing that came to mind). The information-theoretic conception is often used in probability texts as a heuristic for thinking about conditional expectations (for example, see Probability and Measure by Billingsley).
On the information-theoretic conception, we start by thinking of $X$ as a state of uncertainty. To be more concrete, think of $X$ as the unknown outcome of some experiment. A $\sigma$-field $\mathcal X$ on $X$, then, is thought of as "partial information" about the experimental outcome. The idea is: if you "know the information in $\mathcal X$", then, for every $A \in \mathcal X$, you can say whether or not the actual experimental outcome is in $A$.
A function $f$ is measurable on this conception if it is "determined by" or "a function of" the information in $\mathcal X$. How to make this idea precise?
Start with the simplest kind of $\sigma$-field, namely one generated by a finite partition. The obvious way of formalizing the assertion "$f$ is determined by the information in $\mathcal X$" is then:
$f$ is constant on the cells of the partition that generated $\mathcal X$. In other words, if $\mathcal P$ is the finite partition that generates $\mathcal X$, we say that $f$ is measurable in the information-theoretic sense just in case for all $A \in \mathcal P$ and all $x,y \in A$, $f(x)=f(y)$.
There is a straightforward way of generalizing this conception of measurability to arbitrary $\mathcal X$. First, call $a$ an atom of $\mathcal X$ if for some $x \in X$, $a = \bigcap_{x \in A \in \mathcal X}A$. Note that atoms of $\mathcal X$ are not generally members of $\mathcal X$ because they are defined using intersections that are potentially uncountable. Note also that if $\mathcal X$ is generated by a finite partition, then the atoms of $\mathcal X$ are precisely the cells of the generating partition. Generally, then, we have the following definition.
$f$ is measurable in the information-theoretic sense just in case it is constant on the atoms of $\mathcal X$, i.e. for all atoms $a$ of $\mathcal X$ and all $x,y \in a$, $f(x)=f(y)$.
My main question is:
What happens to measure theory if we use the information-theoretic conception of measurability instead of the usual measure-theoretic one? Are there any advantages to doing this? Or do certain important results break down?
Also:
If $\mathcal X$ is countably generated, then every atom of $\mathcal X$ is a member of $\mathcal X$. In this case, I believe the two conceptions of measurability coincide. Is that right?
If so, in order to tease apart the two conceptions of measurability, we will need to consider measurable spaces that aren't countably generated.
Let $\left(\mathbb{R},\mathcal{B}(\mathbb{R})\right)$ be the standard (Borel) measurable space. Then it is countably generated (by family of open intervals with rational endpoints). Moreover, the family of its atoms consists of all singletons of $\mathbb{R}$ and they are members of $\mathcal{B}(\mathbb{R})$.
This implies that the class of information-theoretic measurable functions on $\left(\mathbb{R},\mathcal{B}(\mathbb{R})\right)$ coincides with the class $\mathrm{Map}(\mathbb{R},\mathbb{R})$ of all functions. It also provides negative answer for your second question.