Consider an integrable function $\Omega$ on $S^{n-1}$ with mean value zero, ie $\int_{S^{n-1}}\Omega =0$, and define the tempered distribution
$$\langle W_\Omega, \phi \rangle = \lim_{\varepsilon \to 0} \int_{|x|\geq \varepsilon}\frac{\Omega(x/|x|)}{|x|^n} \phi(x) dx, \;\;\; \forall \phi \in \mathcal{S}{(\mathbb{R}^n)} \;\;\;\;\;\;\;\;(*)$$
The singular integral operator $T_\Omega$ is defined to be the convolution with $W_\Omega$, ie $$T_\Omega \phi = \phi \ast W_\Omega, \;\;\; \forall \phi \in \mathcal{S}{(\mathbb{R}^n)}. $$
It was not a problem for me to verify that $W_\Omega$ is in fact a tempered distribution and so is well defined the singular integral oepator $T_\Omega$. My problem here is about the existence of the limit above. Once proved that $W_\Omega$ is a tempered distribution, is it immediate that the limit exists or I have to show that the limit exists by the Dominated Convergence Theorem?
Attempt: Rewrite the integral $$\int_{|x|\geq \varepsilon}\frac{\Omega(x/|x|)}{|x|^n} \phi(x) dx = \int_{|x|\geq \varepsilon}\frac{\Omega(x/|x|)}{|x|^n} (\phi(x) - \phi(0)) dx.$$ That's possible, because the fact that $\Omega$ has mean value zero implies that $\frac{\Omega(x/|x|)}{|x|^n} \phi(x)$ also has mean value zero. So it is suffices to show that exists the limit $$\lim_{\varepsilon \to 0} \int_{|x|\geq \varepsilon}\frac{\Omega(x/|x|)}{|x|^n} ( \phi(x) - \phi(0)) dx \;\;\;\;\;\;\; (**)$$
By the Mean Value Theorem for $\phi$ \begin{align*} \left|\frac{\Omega(x/|x|)}{|x|^n} ( \phi(x) - \phi(0)) \right| & = \left|\frac{\Omega(x/|x|)}{|x|^n} \nabla\phi(\xi)\cdot x \right|\\ & \leq \frac{|\Omega(x/|x|)|}{|x|^{n-1}} \|\nabla\phi\|_{L^\infty}. \end{align*} By polar coordinates and using the fact that $\Omega \in L^1(S^{n-1})$, its not difficult to show that the function on the right-hand of the inequality is integrable. So by the Dominated Convergence Theorem, the limit $(**)$ exists. So its proved that $(*)$ also exists.