An integral that has a singularity

Question

Question: Does the following integral hold for almost all $x$, where $f$ is a positive measurable function: $$\int_\mathbb{R} \frac{f(t)}{(x - t)^2} dt = +\infty$$

Answer 1

Note that if the integral is finite for even a single $x$, then $f$ is locally integrable on $\mathbb{R}$. Suppose that this is the case. If $x$ is a Lebesgue point of $f$, then $$\lim_{\varepsilon\to 0}\frac{1}{2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon}f(t)dt=f(x).$$ Since $\frac{1}{(x-t)^2}>\frac{1}{\varepsilon^2}$ on the interval $(x-\varepsilon,x+\varepsilon)$, this implies that $$\frac{1}{2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon}\frac{f(t)}{(x-t)^2}dt\gt \frac{f(x)}{2\varepsilon^2}$$ for sufficiently small $\varepsilon>0$, which in turn implies that $$\int_{\mathbb R}\frac{f(t)}{(x-t)^2}dt>\frac{f(x)}{\varepsilon}$$ for sufficiently small, and hence all, $\varepsilon>0$. Thus the integral is infinite for such $x$, and so the integral is infinite almost everywhere by the Lebesgue differentiation theorem.