I was trying to prove the following result:
Define the Hilbert transformation for the continuous function with compact support on $\Bbb{R}$ as $$H f(x):=p \cdot v \cdot \frac{1}{\pi} \int_{\mathbf{R}} \frac{f(y)}{x-y} d y=\lim _{\varepsilon \rightarrow 0} \frac{1}{\pi} \int_{|x-y|>\varepsilon} \frac{f(y)}{x-y} d y$$
Prove the limit above is well defined for continuous function with compact support.
My attempt:First since $f$ is compact supported,only need to consider the integral at a bounded region,what we need to consider is the sigularility at point $x$,I found if we take $f$ be constant near $x$,the integral goes to infinity for one side but the integral has two part,one with negative another one with positive,and we know the integrand change a little bit near $x$ by continuity of $f$,hence there is a hope they cancel out.
It is not true.
Take $f(x) = -\frac{\mathrm{sgn}(x)}{\ln(|x|)}$ near $0$ and $f(0)=0$ and multiply it by a nonnegative smooth function supported on $[-1/2,1/2]$ and with value $1$ in $[-1/4,1/4]$ to make it a compactly supported continuous function.
Then $Hf(0) > \lim_{\varepsilon\to 0}\int_{\varepsilon}^{1/4} \frac{\mathrm d t}{t\,\left|\ln t\right|} = C\,\big(\ln(-\ln(\varepsilon)))-\ln(-\ln(1/4))\big) \to \infty$.
Notice however that by classical results (see e.g. "Fourier Analysis" by Javier Duoandikoetxea) the Hilbert transform is a bounded operator on $L^p$ for any $p\in(1,\infty)$ and so is defined almost everywhere. The simplest case is the fact that $H$ is a bounded operator on $L^2$ since the Fourier transform (in the sense of distributions) is given by $\widehat{Hf}(x) = -i\,\mathrm{sgn}(x)\,\widehat{f}(x)$ and so $\|\widehat{Hf}\|_{L^2} = \|f\|_{L^2}$.