TITLE: Integration with respect to a measure
QUESTION [21 upvotes]: I am trying to get an explanation in words, or math, of what the $d\mu$ means in an integration statement. Such as:
$$\int f \ d\mu$$ 
How does the measure change our old "calculus" notion of integration? What is going on here that is different?

REPLY [25 votes]: I actually struggled with this concept in grad school since I was studying applied math and was sort of thrust into higher level theory without building it up rigorously like I assume would be done in a pure math program.
If we are integrating over a space $X$, I sometimes prefer the notation $\int_X f(x)\mu(dx)$. I like to think of it as splitting the space we are integrating over into infinitesimal pieces, but we have to take the measure of those infinitesimal pieces as they may not all be identical under $\mu$. I'm used to working with probability measures, and at least in that case, you can often think of it similar to the way Lebesgue and Riemann integration are developed.
Create a disjoint partition $X=\cup_{k=1}^N A_k$, and define the sum which will approximate the integral using appropriately chosen sample points $x_k\in A_k$.
$$\int_X f(x)\mu(dx) \approx \sum_{k=1}^N f(x_k) \mu(A_k).$$
Ideally, taking the limit as $N\rightarrow\infty$ (carefully refining the partition and choosing appropriate sample points as $N$ increases) will make the sum converge to the integral. What we are doing here is effectively approximating the function $f$ with a simple function which is constant on a finite collection of measurable sets.