I am trying to show that the following limit exists a.e. for $x\in \mathbb{R}^n$:
\begin{equation}\lim_{\epsilon\to 0}I_\epsilon(x):=\lim_{\epsilon\to 0}\int_{\epsilon<|x-y|<1}\frac{x_j-y_j}{|x-y|^{n+1}}dy,\tag{1}\end{equation} where $x_j$ stands for the $j$-th component of $x$.
This is an exercise from a course I'm following in Harmonic Analysis, and arises in the context of Calderon-Zygmund operators. The integrand is the kernel of the Riesz transforms. We can define the truncated Riesz tranforms as $$R^j_\epsilon(f) (x):=\int_{\epsilon<|x-y|<1}\frac{x_j-y_j}{|x-y|^{n+1}}f(y)dy,$$ and the a.e. convergence of $I_\epsilon(x)$ would imply that the limit $$\lim_{\epsilon\to 0}R^j_\epsilon(f)$$ exists a.e.
Anyway, for $n=1$, I just do \begin{align} I_\epsilon(x)=\int_{\epsilon<|x-y|<1}\frac{x-y}{|x-y|^{2}}dy=\int_{\epsilon<x-y<1}\frac{1}{x-y}dy+\int_{\epsilon<x-y<1}\frac{-1}{x-y}dy=0. \end{align} This is for all $\epsilon>0$, so the limit will be $0$ a.e.. I still doubt whether there's any nuance there, so I'm happy to hear any complaints on the argument.
For $n=2$, I used polar coordinates centered at $x$ to exploit again the antisymmetry of the integrand w.r.t. the hyperplane $y_j=x_j$. Assuming $j=1$ w.l.o.g.:
\begin{align} I_\epsilon(x)=\int_{\epsilon<r<1}\frac{1}{r}\left(\int_{0\leq\varphi<2\pi}\cos\varphi \;d\varphi\right)\,dr, \end{align} which would be again zero for all $\epsilon>0$.
I wonder if this can easily be done in higher dimensions. I am pretty sure that it works in $n=3$, but I don't know how to express the integral in hyperspherical coordinates so that the symmetric property of the integrand shows up meaningfully. I would appreciate either a hint on this, or another way to tackle the proof.