$\int_{|y|>\epsilon}\frac{|\Omega(y/|y|)|}{|y|^n}|f(x-y)|\,dy<\infty \text{ for a.e.} x\in \mathbb R^n$, where $f\in L^p$ and $\Omega\in L^1(S^{n-1})$

Question

Let $\Omega\in L^1(S^{n-1})$, where $S^{n-1}$ is the unit sphere in $\mathbb R^n$. Let $f\in L^p(\mathbb R^n)$ for some $p\in[1,\infty)$. For each $\epsilon>0$, prove that $$\int_{|y|>\epsilon}\frac{|\Omega(y')|}{|y|^n}|f(x-y)|\,dy<\infty ,\qquad\text{a.e. } x\in \mathbb R^n.$$ Here $\Omega(y')=\Omega(y/|y|)$.

I can only prove the result for $p\in[1,\frac{n}{n-1})$.

This problem comes from the teacher of my Harmonic Analysis course. I've tried to prove $$\int_{|x|\leq R}\int_{|y|>\epsilon}\frac{|\Omega(y')|}{|y|^n}|f(x-y)|\,dy\,dx<\infty$$ for each $R>0$. Note: In the following I'll use $\chi_A$ to denote the indicate function of the set $A$.

\begin{align*} &\qquad\int_{|x|\leq R}\int_{|y|>\epsilon}\frac{|\Omega(y')|}{|y|^n}|f(x-y)|\,dy\,dx\\ &=\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{|\Omega(y')|}{|y|^n}|f(x-y)|\chi_{|x|\leq R}\ \ \chi_{|y|>\epsilon}\,dx\,dy\\ &=\int_{\mathbb R^n}\int_{\mathbb R^n}\frac{|\Omega(y')|}{|y|^n}|f(z)|\chi_{|y+z|\leq R}\ \ \chi_{|y|>\epsilon}\,dy\,dz\\ &=\int_{\mathbb R^n}|f(z)|\left(\int_{\mathbb R^n}\frac{|\Omega(y')|}{|y|^n}\chi_{|y+z|\leq R}\ \ \chi_{|y|>\epsilon}\,dy\right)\,dz \end{align*}

On the one hand, for each fixed $z\in\mathbb R^n$, we have $$\int_{\mathbb R^n}\frac{|\Omega(y')|}{|y|^n}\chi_{|y+z|\leq R}\ \ \chi_{|y|>\epsilon}\,dy\leq\int_{\epsilon<|y|\leq R+|z|}\frac{|\Omega(y')|}{|y|^n}\,dy\leq C\|\Omega\|_{L^1(S^{n-1})}\ln\frac{|z|+R}\epsilon.$$ This is sufficient for estimating the integral on $\{|z|\leq 2R\}$.

On the other hand, for those $z$ with $|z|\geq 2R$, we have $$\int_{\mathbb R^n}\frac{|\Omega(y')|}{|y|^n}\chi_{|y+z|\leq R}\ \ \chi_{|y|>\epsilon}\,dy\leq\int_{|z|-R<|y|\leq R+|z|}\frac{|\Omega(y')|}{|y|^n}\,dy\leq C\|\Omega\|_{L^1(S^{n-1})}\ln\frac{|z|+R}{|z|-R}.$$ To ensure $$\int_{|z|>2R}\ln\frac{|z|+R}{|z|-R}|f(z)|\,dz<\infty,$$ it suffices that $\ln\frac{|z|+R}{|z|-R}\in L^{p'}(|z|>2R)$, where $p'$ is the conjugate exponent of $p$, and this is equivalent to $p'>n$, or $p<\frac{n}{n-1}$.

However, I don't know how to prove the result for other $p$'s.

Any help would be appreciated.