= Solution
Define the joint <function> by an unambiguous piecewise prescription:
$$
H(x,y)=\begin{cases}(x-y)^{\beta-1}f(y),&0<y<x<1,\\0,&\text{otherwise in }(0,1)^2.\end{cases}
$$
It is nonnegative and <Lebesgue measurable>: on the open triangle the power factor is continuous and $f(y)$ is measurable for the <product sigma-algebra>, and the zero extension is measurable. This also resolves the diagonal singularity and avoids evaluating a noninteger power of a negative number outside the triangle. The source's inclusion of $y=x$ in its indicator <set> changes only a <null set>; set the value there to zero.
The <function> $g(x)=\int_0^1H(x,y)\,dy$ is measurable by <Tonelli's theorem>. Apply that theorem to exchange the nonnegative <integrals>, with infinity allowed until the estimate proves otherwise:
$$
\begin{aligned}
\int_0^1g(x)\,dx
&=\int_0^1f(y)\left(\int_y^1(x-y)^{\beta-1}\,dx\right)dy\\
&=\frac1\beta\int_0^1(1-y)^\beta f(y)\,dy\\
&\leq\frac1\beta\int_0^1f(y)\,dy<\infty.
\end{aligned}
$$
The inner <integral> is $(1-y)^\beta/\beta$ because $\beta>0$. The weight is at most one because $0<y<1$. Since $g\geq0$, this proves
$$
\boxed{\|g\|_1=\frac1\beta\int_0^1(1-y)^\beta f(y)\,dy
\leq\frac1\beta\|f\|_1,\qquad g\in L^1((0,1)).}
$$
In particular $g$ is finite <almost everywhere>; any exceptional infinite values may be replaced by zero to obtain an everywhere finite representative. This is the <L1 bound for a Volterra power-kernel integral>.
An everywhere-finite conclusion would be stronger and can fail when $0<\beta<1$, even for integrable $f$. For example, put $x_0=1/2$ and $\gamma=(1+\beta)/2$, and take $f(y)=(x_0-y)^{-\gamma}$ for $0<y<x_0$ and zero otherwise. Since $\gamma<1$, $f$ is integrable. But
$$
g(x_0)=\int_0^{x_0}(x_0-y)^{\beta-1-\gamma}\,dy=\infty,
$$
because $\beta-1-\gamma=(\beta-3)/2<-1$. This does not contradict <Lebesgue integrability>, which requires finiteness only <almost everywhere>.
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