I started a lecture about harmonic analysis and early on the Hilbert Transformation is defined.
$$(Hf)(x) = \frac{1}{\pi}\lim_{\varepsilon\rightarrow 0}\int_{|y|>\varepsilon} \frac{f(x-y)}{y} \operatorname{d}y$$
Assume $f\in\mathcal{C}^1(\mathbb R)\cap L^1(\mathbb R)$. The lecturer notes then looks at the limit term $$t(x,\varepsilon):=\int_{|y|>\varepsilon} \frac{f(x-y)}{y} \operatorname{d}y,$$ notes that since $|y|>\varepsilon$ we have $t(x,\varepsilon)<\frac{|f|_1}{\varepsilon}$ and goes on to show that
$$ \begin{aligned} |t(x,\varepsilon)| &\leq \int_{|y|\geq 1} \frac{|f(x-y)|}{|y|}\operatorname{d}y + \left|\int_{\varepsilon< |y| < 1} \frac{f(x-y)-f(x)}{y}\operatorname{d}y\right| \\ &\leq |f|_1 + |f'|_\infty \int_{\varepsilon< |y| < 1} 1 \operatorname{d}y \\ &\leq |f|_1 + 2|f'|_\infty \end{aligned} $$
thereby showing that we can bound $t(x,\varepsilon)$ independent of $\varepsilon$. Then he notes that $$\left\{t(x,\varepsilon)\right\}_{\varepsilon > 0}$$ can be shown to be a Cauchy sequence using a similar argument, but I can't figure it out.
I would presume we do something like $$ \begin{aligned} |t(x,\varepsilon_2)-t(x,\varepsilon_1)| &\leq \left|\int_{\varepsilon_1 < |y| < \varepsilon_2} \frac{f(x-y)-f(x)}{y}\operatorname{d}y\right| \\ &\leq |f'|_\infty(\varepsilon_2 - \varepsilon_1) \end{aligned} $$ but it can't be that simple, right?