I think this should be true.
Let $f\in L^{1}(\Bbb{R})$ and consider $g(x)=\frac{1}{\sqrt{|x|}}$ , then is $(f*g)\,(y)$ well defined for $y$ almost everywhere?
I know that convolution of two $L^{1}$ functions is well defined but here $\frac{1}{\sqrt{|x|}}$ is not $L^{1}(\Bbb{R})$.
Instead what I was thinking was to consider $$\int_{\Bbb{R}}\frac{f(y-x)}{\sqrt{|x|}}\,dx=\int_{[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx+\int_{\Bbb{R}\setminus[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx$$
Now, for $|x|\geq 1$ we have $\frac{1}{\sqrt{|x|}}\leq 1$ hence $\displaystyle\int_{\Bbb{R}\setminus[-1,1]}\frac{|f(y-x)|}{\sqrt{|x|}}\,dx\leq ||f||_{L^{1}}$.
Hence $\int_{\Bbb{R}\setminus[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx$ is well defined almost everywhere in $\Bbb{R}\setminus [-1,1]$ as the integrals over the positive and negative parts of the above function will be finite.
We have that $\frac{1}{\sqrt{|x|}}$ is $L^{1}([-1,1])$. Hence, By Tonelli's Theorem, we have
$$\int_{[-1,1]^{2}}\frac{|f(y-x)|}{\sqrt{|x|}}\,d(x,y)=\int_{[-1,1]}\int_{[-1,1]}\frac{|f(y-x)|}{\sqrt{|x|}}\,dy\,dx\leq 4\cdot||f||_{L^{1}} $$
where the $4$ comes from $\int_{[-1,1]}\frac{1}{\sqrt{|x|}}\,dx$ and we have used the translational invariance property to conclude that $\int_{[-1,1]}|f(y-x)|\,dy\leq\int_{\Bbb{R}}|f(y-x)|\,dy=||f||_{L^{1}}$.
Hence as $\displaystyle\int_{[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx$ is integrable over $[-1,1]$, it is finite almost surely over $[-1,1]$
Coupling both we have $\int_{[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx+\int_{\Bbb{R}\setminus[-1,1]}\frac{f(y-x)}{\sqrt{|x|}}\,dx=\int_{\Bbb{R}}\frac{f(y-x)}{\sqrt{|x|}}\,dx=(f*g)(y)$ is well defined for $y$ almost everywhere.
Question: Is the above argument correct and is there any other (possibly shorter) way to do this? By shorter I obviously don't mean the omission of the (possibly) verbose justifications I gave at each step. Also, if the above is correct, we can say that the convolution with $\frac{1}{|x|^{a}}$ is well defined when $0<a<1$.
It works for every $g=g_1+g_\infty\in L^1 + L^\infty$. This is essentially the same idea of cutting the function, but a shorter proof is by using Young's inequality. Indeed, it follows from Young's inequality that $$ g_1* f\in L^1 $$ and $$ g_\infty* f\in L^\infty $$ Therefore $g* f\in L^1+L^\infty$ is defined almost everywhere.