Past exam of the integration course of the University of Oxford / 2014 / 2

The dominated convergence theorem states: on a measure space $(X,\mathcal{F},\mu )$, let $u_n$ be measurable functions converging almost everywhere to a measurable $u$. Suppose there is a nonnegative measurable function $G$ with ${\int }_{X}Gd\mu <\infty$ and $|u_n|\le G$ almost everywhere for every $n$. Then $u$ is integrable andHere integrability means integrability with respect to the given measure; for Lebesgue measure it is Lebesgue integrability. Neither a finite measure space nor monotonicity of the sequence is required. The same single integrable dominator must control the whole sequence.

Fix $t_0\ge 0$. On a compact parameter interval containing all nonnegative $t$ sufficiently close to $t_0$, the corresponding rectangle with $x\in [0,1]$ is compact. The continuous function $f$ is therefore bounded there by some $M<\infty$, by the extreme value theorem.

For any sequence $t_n\ge 0$ tending to $t_0$, all sufficiently late $t_n$ lie in this rectangle's parameter range. For each $x\in [0,1]$, continuity givesThe constant function $M$ is integrable on $[0,1]$. The dominated convergence theorem therefore givesThe sequential criterion for continuity on a real interval proves $F$ is continuous on $[0,\infty )$. At zero this is right-continuity. This is continuity of parameter integrals on compact rectangles; the bound is local in the parameter, so $f$ need not be bounded on the entire unbounded strip.

We prove the required differentiation under the integral sign directly. Fix $t_0>0$, and take a compact parameter interval $J\subseteq [0,\infty )$ containing $t_0$ in its interior. The jointly continuous partial derivative ${\partial }_{t}f$ is bounded on the compact rectangle $J\times [0,1]$ by some $M_1$.

For nonzero $h$ sufficiently close to zero, defineFor each $x$, the mean value theorem in the parameter gives $|D_h(x)|\le M_1$. The definition of the partial derivative also gives $D_h(x)\to {\partial }_{t}f(t_0,x)$. Since $M_1$ is integrable on $[0,1]$, dominated convergence applied along any sequence $h_n\to 0$ yieldsThusAt $t=0$, repeat the argument on $J=[0,\delta ]$ with $h>0$. It proves that the right derivative exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact $F^{\prime }$ is continuous there: apply part (i) to the continuous function ${\partial }_{t}f$.

The proof establishes differentiation of parameter integrals on compact rectangles under the exact regularity assumptions used here. In particular, it supplies the theorem's proof instead of assuming an interchange of a limit and an integral.

Remove the apparent singularity at $x=0$ by definingThis is jointly continuous: write $\varphi (t,x)=t\mathrm{sinc}(tx)$ using the unnormalized sinc function, whose value at zero is one. Its partial derivative in $t$ isalso at $x=0$, since there $\varphi (t,0)=t$. This partial derivative is jointly continuous on the whole strip.

The value at the single endpoint $x=0$ does not change a Lebesgue integral. Part (b), whose dominated convergence proof applies to $\varphi$, therefore makes $g$ continuously differentiable on $[0,\infty )$, with a right derivative at zero, andFor $t>0$, direct integration givesAt zero the same integral gives $\boxed{g_{+}^{\prime }(0)=1}$, so no division by zero is required. Also $g(0)=0$. The substitution rule $u=tx$ identifies $g(t)={\int }_0^t\sin u/udu$ for $t>0$, the sine integral.