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

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.

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.