# A set with a finite integral of measure zero?

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.

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Maybe the Lebesgue decomposition theorem can be of some help? If $\mu$ is absolutely continuous or discrete the claim is easy to prove. EDIT: This is the theorem I'm referring to – Giuseppe Negro Apr 3 '11 at 19:18
I thought about it. But I don't see any easy way to solve either. Perhaps the continuous part (in respect to Lebesgue measure) could be shown to satisfy the infinite integral in any Lebesgue point on which the Radon differential has a non-zero value. But how would you deal with the singular part? – yaakov Apr 3 '11 at 19:26
Isn't any discrete measure a sum of Dirac's deltas? I recall something like that. Let it be $\nu=\sum_{n=0}^\infty \lambda_n \delta_{x_n}$: so $\int(x-y)^{-2}\, d\nu=+\infty$ when $x=x_n$, that is, $\nu$-almost everywhere. – Giuseppe Negro Apr 3 '11 at 19:30
why should it be? you can take any set of measure zero, and give it some funny distribution (take the Cantor distribution for example). EDIT: I think you are mistaking "singular" for "discrete", those are separate concepts. – yaakov Apr 3 '11 at 19:32
Oh, you edited. Yes, in fact I was referring to the discrete part, not the singular part. Neither I have any clue on how to handle the latter. – Giuseppe Negro Apr 3 '11 at 19:37

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 $\int\vert x-y\vert^{-1}\,d\mu(y)$.

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Could you please explain why $\mu$ cannot have measure greater than $K\epsilon^2$ on an interval of length $\epsilon$ intersecting $S$? (I know it's been a while since you posted this answer, but I am seeing it only now, after more than four years!) – Giuseppe Negro Sep 11 '15 at 21:05
@GiuseppeNegro: If $I$ is an interval of width $\epsilon$ and $x\in S\cap I$, then, the function $y\mapsto\epsilon^2(x-y)^{-2}$ is bounded below by $1$ on this interval. So, $\mu(I)\le\epsilon^2\int(x-y)^{-2}\mu(dy)\le\epsilon^2K$. – George Lowther Sep 15 '15 at 19:46
I have got it. Thank you. – Giuseppe Negro Sep 16 '15 at 23:32

this is a clumsy solution and certainly too long - but anyway. The answer is yes. There is a (weakly) increasing function $f:I\to\mathbb{R}$ such that $\mu$ is the push-forward of the Lebesgue measure on $I$ to $\mathbb{R}$ ($f$ is, roughly speaking, the inverse of $F(x)=\mu((-\infty,x])$). Your integral becomes $\int_I (f(x)-f(y))^{-2} dy$ (where "your $x$" is "my $f(x)$"). Since $f$ is monotone, it has a derivative a.e. If $x$ is in the set where the derivative is $<A$ (for any positive $A$) then the integral is $\infty$ by comparison with $\int_I (x-y)^{-2} dy$. Hence the integral is $\infty$ for a.e. $x$.

(not very pretty - but possibly correct :)

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Looks like it works. And it should also apply to $\int\vert x-y\vert^{-1}\,d\mu(y)$. Or, indeed, to $\int\theta(\vert x-y\vert)\,dy$ for any decreasing function $\theta$ satisfying $\int^{0+}\theta(x)\,dx=\infty$. – George Lowther Apr 3 '11 at 22:30