We start our turn from measure in the abstract to applying it to integration, and we start with simple functions. In fact, we start a bit further back than that even; the simple functions are exactly the finite linear combinations of characteristic functions, and so we’ll start there.
Given a measurable set , there’s an obvious way to define the integral of the characteristic function : the measure ! In fact, if you go back to the “area under the curve” definition of the Riemann integral, this makes sense: the graph of is a “rectangle” (possibly in many pieces) with one side a line of length and the other “side” the set . Since is our notion of the “size” of , the “area” will be the product of and . And so we define
That is, the integral of the characteristic function with respect to the measure is . Of course, this only really makes sense if .
Now, we’re going to want our integral to be linear, and so given a linear combination we define the integral
Again, this only really makes sense if all the associated to nonzero have finite measure. When this happens, we call our function “integrable”.
Since every simple function is a finite linear combination of characteristic functions, we can always use this to define the integral of any simple function. But there might be a problem: what if we have two different representations of a simple function as linear combinations of characteristic functions? Do we always get the same integral?
Well, first off we can always choose an expression for so that the are disjoint. As an example, say that we write , where and overlap. We can rewrite this as . If is integrable, then and both have finite measure, and so is subtractive. Thus we can verify
Thus given any representation the corresponding disjoint representation gives us the same integral.
But what if we have two different disjoint representations and ? Our function can only take a finite number of nonzero values . We can define to be the (measurable) set where takes the value . For any given , we can consider all the so that . The corresponding sets must be a disjoint partition of , and additivity tells us that the sum of these is equal to . But the same goes for the corresponding to values . And so both our representations give the same integral as . Everything in sight is linear, so this is all very straightforward.
At the end of the day, the integral of any simple function is well-defined so long as all the preimage of each nonzero value has a finite measure. Again, we call these simple functions “integrable”.