Why do we consider measurable function when dealing with abstract integration? Let $[0,\infty]$ be equipped with the order topology. (That is, it is a subspace of the standard topology on the extended real)
Let $(X,\mathfrak{M},\mu)$ be a measure space.
Let $f:X\rightarrow [0,\infty]$ be a measurable function, which means that the inverse of each borel set of $[0,\infty]$ under $f$ is an element of $\mathfrak{M}$.
In measure theory, we define it's integration as the supremum of $\sum_{x\in s(X)} x\mu(s^{-1}(x))$ where $s$ is a simple measurable fuction such that $s≦f$.
The point is, the role of $\mu$ only acts on those simple functions $s$, not $f$ directly.
What would go wrong if one defines integration for an arbitrary function $f$?
That is, what's wrong with defining an integration of an arbitrary function $f$ as the supremum of $\sum_{x\in s(X)} x\mu(s^{-1}(x))$ over simple meadurable functions $s≦f$?
 A: Given a Vitali $E$ set in $[0, 1]$, there are no measurable subsets of $E$ with positive measure (see, e.g. here). Hence if we want to integrate $f = \chi_E$, then we would necessarily have $\int f dm = 0$. But one of the "nice" properties of Lebesgue integration is that any nonnegative function with zero integral must be zero almost everywhere - but this implies $E$ has measure $0$, contradicting that it isn't measurable.
So all of a sudden, a lot of results that rely on the idea of equality almost everywhere, or a property holding up to sets of measure $0$ would break down. 

Furthermore, the idea that integration and measure are related would break down fundamentally, were this the case. We'd really like to have that
$$\int \chi_A d\mu = \mu(A)$$
for measures; in fact, it's a general result that the function $A \mapsto \int \chi_A f d\mu$ is a measure for measurable functions $f$ (or taking $f$ identically $1$, giving back the measure $\mu$). But if every function is integrable, every set is measurable; but having regularity of the measure requires that some sets aren't measurable.
