 Solution

Define the joint function by an unambiguous piecewise prescription: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 $g(x)={\int }_0^{1}H(x,y)dy$ is measurable by Tonelli's theorem. Apply that theorem to exchange the nonnegative integrals, with infinity allowed until the estimate proves otherwise:The inner integral is $(1-y{)}^{\beta }/\beta$ because $\beta >0$. The weight is at most one because $0<y<1$. Since $g\ge 0$, this provesIn 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. Butbecause $\beta -1-\gamma =(\beta -3)/2<-1$. This does not contradict Lebesgue integrability, which requires finiteness only almost everywhere.