TITLE: A set with a finite integral of measure zero?
QUESTION [11 upvotes]: Prove, or give a counter example:
Let $\mu$ be a finite positive borel measure on $\mathbb{R}$. Then $\int (x-y)^{-2} d \mu (y) = \infty $ almost everywhere on $\mu$ (for the selection of x's).
This is a question I had in an exam, and the answer is supposed to be presented in less than 30 words, so there must be something quite simple I'm missing.

REPLY [5 votes]: Yes it is true. Consider the set $S$ on which the integral in question is bounded by a positive value $K$. Then $\mu$ cannot have measure greater than $K\epsilon^2$ on any interval of width $\epsilon$ intersecting $S$. However, any interval $[a,b]$ can be divided into $n$ intervals of width $(b-a)/n$. So, $\mu(S\cap[a,b])$ is bounded by $n K ((b-a)/n)^2$. Let $n$ go to infinity.
I didn't count, but I know that exceeded 30 words by a fair margin.
The argument is easily adapted to show that $\int\vert x-y\vert^{-1-\epsilon}\,d\mu(y)$ is infinite $\mu$-almost everywhere for any $\epsilon > 0$. The more difficult question is whether the same holds for $\int\vert x-y\vert^{-1}\,d\mu(y)$.