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

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be measure spaces with sigma-finite measures, and use their product measure.

Fubini's theorem: if $u$ is a measurable absolutely integrable function on $X\times Y$, then its sections are integrable for almost every fixed value of the other coordinate, the resulting section-integral functions are integrable, andAll three integrals are finite. Values of an inner integral on an exceptional null set can be assigned arbitrarily.

Tonelli's theorem: if $u:X\times Y\to [0,\infty ]$ is measurable and nonnegative, the same equality holds for its nonnegative product and iterated integrals, allowing the common value $+\infty$. No advance absolute integrability assumption is needed. On a completed Lebesgue measure product, the section and measurability assertions are interpreted almost everywhere.

Thus Tonelli's theorem can first establish that $\int |u|<\infty$; after this, Fubini's theorem permits signed integration in either order.

Assign any value, for example zero, at $(0,0)$, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

In polar coordinates, $x^2+y^2=r^2$ and the area element is $rdrd\theta$. Applying the nonnegative change of variables formula to the absolute value givesFor $0<r\le 1$, $|\sin (r^2)|\le r^2$, so the radial integrand is bounded by $r^{3-2\alpha }$. This is integrable at zero when $\alpha <2$. For $r\ge 1$, $|\sin (r^2)|\le 1$, giving the integrable tail bound $r^{1-2\alpha }$ when $\alpha >1$. In particular,ThereforeEquivalently, the substitution rule $u=r^2$ reduces its exact L1 norm toThis is radial sine-power kernel integrability, reducing a two-dimensional question to the sine power kernel integrability thresholds at zero and infinity.

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.