Showing that the integral with respect to an abstract measure is infinite Suppose $\mu$ is a measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. Then show that, for $\mu$-almost every $x$
$$
 \int_\mathbb{R} \frac{1}{(x-t)^2}\,\mathrm{d}\mu(t) = \infty
$$
I am not sure where to start. Any hints are welcome.
 A: Technicality: In this sort of context "measure" usually means "regular Borel mmeasure". We need to  assume that $\mu$ is a Borel measure or there's no reason that the integrand should be measurable. We don't need to worry about regularity: All we need from regularity below is that $\mu(K)<\infty$ for compact $K$, and this is something we can assume wlog, because if $K$ is compact and $\mu(K)=\infty$ then the integral is infinite for every $x$.
This is curious. If the integral is not infinite for almost every $x$ then by countable additivity there exist $C<\infty$ and a set $E$ with $\mu(E)>0$ such that $$\int_{\mathbb R}\frac1{(t-x)^2}d\mu(t)\le C\quad(x\in E).$$
Again by countable additivity there exists $a$ so that $\mu(F)>0$, if $$F=E\cap[a,a+1).$$ Note that $\mu(F)<\infty$, by the "technicality" above.
The inequality above shows that $$\int_F\int_F\frac1{(t-x)^2}d\mu(t)d\mu(x)\le C\mu(F).$$
Now fix $n$ for the moment and let $$F_j=F\cap[a+j/n,a+(j+1)/n)\quad(0\le j<n).$$
Then $$\int\int_{F_j\times F_j}\frac1{(t-x)^2}d\mu(t)d\mu(x)\ge n^2\mu(F_j)^2.$$ Since the $F_j\times F_j$ are disjoint subsets of $F\times F$ this shows that $$n^2\sum\mu(F_j)^2\le C\mu(F).$$
But since $\sum\mu(F_j)=\mu(F)$, Cauchy-Schwarz shows that $$\mu(F)^2\le n\sum\mu(F_j)^2.$$Combining two inequalities above now gives $$n\mu(F)^2\le C\mu(F).$$
Now let $n\to\infty$ and recall that $0<\mu(F)<\infty$.
