Integrating when the boundaries converge to each other Let's have a look at a fictitious matriarchal society in which women pick multiple spouses. 
Attractiveness is indexed between 0 and 1. Let $m(x)$ denote the mass of males with attractiveness $x$. For each value of $x$ on the unit interval, we also have a measure of females with that attractiveness. These females will take as spouse anyone who is at least as attractive as them. 
For a given $x$, we can compute this pool as 
$$ f(x) = \int_x^1 m(i) di $$
I'm wondering what happens if we look at $f(1)$. From the equation, it should be that $f(1) = \int_1^1 m(i) di = 0$ for any $m(x)$. Even if I don't look at $f(1)$ but $\lim_{x\to 1} f(x)$, it still appears that I get the same result.
However, if I go back to my story, I "feel" that the pool should be $m(1)$ - which is the set of people at least as good as 1.
How can I resolve this?
 A: First, let $m(x)=M'(x)$, so that $\int_x^1 M'(i)di=M(1)-M(x)$. From this we see that as $x\rightarrow 1$ the integral vanishes.
As for the interpretation of why it vanishes, I think your point of trouble comes from the type of function $m(x)$ is. The problem is that where $m$ is defined ([0,1]), it is not a discrete function. Since you are assuming it is integrable it should be at least piecewise continuous. I don't know if you've taken any probability lectures, but the same happens for example in a probability distribution: the probability that $x$ lies in between 0 and 0.00001 could be different than zero, but the probability that $x=0$ is $0$, because you are integrating over nothing.
Specifically in this example, we would have the most attractive woman in the tribe ranked 1 in attractiveness, and she would only chose men ranked 1 as well. So (assuming it doesn't 'blow up' at 1) she only has a finite number of men to chose from. Now if you picture the typical introduction to integrals, they are a sum of infinitely many 'bins' or rectangles that have the height of a function at a point. The problem is the 'width' of the number 1, is 0. Intervals have width, but numbers don't; and so your 'bin' is of height $m(1)$, but of width $0$, and therefore contributes $0$ to the integral. 
Since this is a 'problem' inherent to the integral, you could step around it by defining $f(x)$ as a split function, it being the integral where $x\in[0,1)$, and $f(x)=m(1)$ when $x=1$.
