Past exam of the integration course of the University of Oxford / 2014 / 3 / c / i / Solution

Fix $0<x<1$. On the open interval $(0,x)$, the function $y\mapsto (x-y{)}^{\beta -1}$ is finite, positive and continuous, and $f$ is nonnegative and Lebesgue measurable. Their product is therefore a nonnegative measurable function. Its Lebesgue integral is always defined as an element of $[0,\infty ]$, with no possibility of cancellation of two infinities. Thus $g(x)$ is well defined as a nonnegative extended integral.

The endpoint $y=x$ need not receive its formal power value: for $\beta <1$ that value would be singular. The integral is over $(0,x)$, and any chosen value at the endpoints has no effect because endpoints form a null set.

Finiteness is not implied by the present assumptions. For example, $f(y)=1/y$ is an admissible nonnegative measurable function. For any fixed $x$, the factor $(x-y{)}^{\beta -1}$ is bounded below by a positive constant on $0<y<x/2$, so $g(x)\ge C_x{\int }_0^{x/2}dy/y=\infty$. The later L1 norm bound will establish finiteness almost everywhere when $f$ is integrable. The construction is a Volterra power-kernel integral.