Is $g(x)=\int_a^x (x-s)^{-\alpha}f(s) \, ds\in AC[a,b]$ if $f\in L_1[a,b]$ and $\alpha\in (0,1)$?

Question

Let $f\in L_1[a,b]$ and $\alpha\in(0,1).$ Is it true that the function $$ g(x)=\int_a^x \frac{f(s)}{(x-s)^\alpha}\,ds $$ is absolutely continuous on $[a,b]$? First, I thought it is true and I tried to show that there exists $\phi\in L_{1}[a,b]$ such that $$g(x)-g(y)=\int_y^x \phi(s) \, ds$$ but it was very difficult (for me). Then I tried to find a function $f\in L_1 [a,b]$ such that $g\not\in AC[a,b]$ but I couldn't. Any idea please?

Answer 1

Consider $\alpha = 1/2$, $f(s)=(x-s)^{-1/2}$. Then the integral does not even exist.