An integral that has a singularity

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

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Yes, it's divergent, but some meaning could still be attributed to it... –  Ｊ. Ｍ. Jun 2 '11 at 5:40
Thank you. But I'm unable to produce a proof for this unless f is nice, say, being continuous, etc. –  Ken Leung Jun 2 '11 at 5:55
–  user8268 Jun 2 '11 at 7:16

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.