In "Singular Integrals and Differentiability Properties of Functions" by Stein he asserts on page 40 that
\begin{equation} \int_{|x|\geq 2|y|}\frac{\Omega(x-y)-\Omega(x)}{|x-y|^n} dx \leq C \quad (1) \end{equation}
His argument goes as follows: Since $\left| \frac{x-y}{|x-y|}-\frac{x}{|x|} \right| \leq C \left|\frac{y}{x}\right|$ for $|x|\geq 2|y|$ (first of all, how is the last fraction defined, AFAIK $x$ and $y$ are vectors so their quotient is ill-defined? The only thing I can make sense of is $\frac{|y|}{|x|}$). Now Stein assumes some condition on $\Omega$, namely $$\sup_{|x-x'|\leq \delta\\ |x|=|x'|=1} \Omega(x)-\Omega(x') = \omega(\delta) \quad \text{implies} \quad \int_0^1 \frac{\omega(\delta) d\delta}{\delta} < \infty$$ First of all the indices of $\sup$ are barely readable, so I don't know if I even copied them correctly! Together with this condition we are led to believe that "the integral corresponding to the first group of terms (i.e. (1)) is dominated by" $$ C' \int_{|x|\geq 2|y|} \omega\left(C\frac{|y|}{|x|}\right) \frac{dx}{|x|^n} = C''\int_{0}^{c/2} \frac{\omega(\delta)}{\delta} d\delta < \infty$$
Unfortunately I have no clue how this all works out. How do we get to the last line from the conditions listed above?
Ok I think I figured this out. First of all observe that $$\frac{1}{|x-y|^n} \leq \frac{2^n}{|x|^n}$$ since $|x|\geq 2|y|$ (which implies $|x-y| \geq |x|/2$ by the reverse triangle inequality). Then we have that (and I didn't mention that in my question) that $\Omega$ is homogeneous of degree zero, i.e. $\Omega(\epsilon x) = \Omega(x)$ for all $\epsilon > 0$. Hence $$\Omega(x-y)-\Omega(x) = \Omega\left(\frac{x-y}{|x-y|}\right)-\Omega\left(\frac{x}{|x|}\right) .$$ Moreover, the condition $\left| \frac{x-y}{|x-y|}-\frac{x}{|x|} \right| \leq C \left|\frac{y}{x}\right|$ holds for all $x,y$ so it certainly holds that $$\Omega(x-y)-\Omega(x) \leq \sup_{\left| \frac{x-y}{|x-y|}-\frac{x}{|x|} \right| \leq C \left|\frac{y}{x}\right|} \Omega\left(\frac{x-y}{|x-y|}\right)-\Omega\left(\frac{x}{|x|}\right) = \omega\left(C\frac{|y|}{|x|}\right)$$ Putting it all together yields $$\int_{|x|\geq 2|y|}\frac{\Omega(x-y)-\Omega(x)}{|x-y|^n} dx \leq 2^n \int_{|x|\geq 2|y|} \omega\left(C\frac{|y|}{|x|}\right) \frac{dx}{|x|^n}$$ and by a change of variables $x = |y|z$ and introducing polar coordinates we can exploit the integrability condition $\int_0^1 \frac{\omega(\delta) d\delta}{\delta} < \infty$.