Let $0<a<1$ and $f\in L^1([0,1])$. Show $g(x)=\int_0 ^x\frac{1}{(x-t)^a}f(t)dt$ exists a.e. in $[0,1]$ and $g\in L^1([0,1])$.
Using Fubini, $$\int_0 ^1 \vert g(x) \vert dx=\int_0 ^1 \int_0 ^x\frac{f(t)}{(x-t)^a}dtdx=\\\int_0 ^x \bigg( \int_0 ^1\frac{1}{(x-t)^a}dx\bigg)dt=\int_0 ^x\frac{1}{(1-a)(x-t)^{a-1}}\bigg\vert_0 ^1dt\\=\int_0 ^x\bigg(\frac{1}{(1-a)(1-t)}-\frac{1}{(1-a)(-t)^{a-1}}\bigg)dt\\=\bigg[-\log(1-t)/(1-a)- \frac{1}{(1-a)(2-a)(-t)^{a-2}}\bigg]\bigg\vert _0 ^x\\=-\log(1-x)/(1-a)-\frac{1}{(1-a)(2-a)(-x)^{a-2}}$$
I beleive this shows that $g$ is in $L^1$, but what's needed to show it exists a.e.?
Update My approach is flawed and I don't have an idea of how to proceed.
Let $f:X \rightarrow [0, +\infty]$ be a measurable function and $A$ a measurable subset of $X$. If measure of a set $C = \{x \in A : f(x) = +\infty\}$ is strictly positive, then $\int_A \! f \, \mathrm{d}\mu = +\infty.$
Contraposition gives us that for $g$ to exist almost everywhere, it is enough that $|g|$ is integrable. That is why it is enough to prove that $g \in L^{1}$.
As for a proof of that, your idea is good, just don't forget $|f|$ - because $f\in L^{1}$, it is easy to bound $||g||$ the way you did.