Past exam of the integration course of the University of Oxford / 2021 (A4, A13756W1)

Source: original released examination PDF, bundle pages 79–81; paper pages 1–3.

Solutions to all three questions of the examination sat on 16 June 2021, including all optional parts. Integrability means Lebesgue integrability; a conditionally convergent improper integral does not suffice. The Roman items in Question 3(a) are the three required properties of one construction; the construction and its limiting properties are explained first, and each Roman item is checked separately.

Extend the integrand by zero to all of $(0,\infty )$:For each fixed $x>0$, eventually the indicator function is one and $\sin (x/n)/(x/n)\to 1$. Hence $h_n(x)\to (1+x^2{)}^{-1}$ pointwise. On its support, $x/n\le 2\pi$, so $|\sin t|\le t$ for $t\ge 0$ givesThe final bound is Lebesgue integrable on $(0,\infty )$ and independent of $n$. The dominated convergence theorem therefore givesThe last integral follows from the antiderivative $\arctan x$.

On $(0,1)$, the integrand is bounded by one. For $x\to \infty$, $\alpha x^{\beta }-2\log x\to \infty$, since every positive power grows faster than the natural logarithm. Thus $e^{-\alpha x^{\beta }}\le x^{-2}$ eventually. The comparison test for integrals proves Lebesgue integrability.

To prove joint continuity, fix $({\alpha }_0,{\beta }_0)$ with both coordinates positive. In a sufficiently small parameter rectangle, $\alpha \ge a:={\alpha }_0/2>0$ and $\beta \ge b:={\beta }_0/2>0$. For every parameter pair in this rectangle,The first argument shows that $H$ is integrable. Along any parameter sequence tending to $({\alpha }_0,{\beta }_0)$, the integrands converge pointwise, and the same $H$ dominates their eventual tail. The dominated convergence theorem gives $g(\alpha ,\beta )\to g({\alpha }_0,{\beta }_0)$. The sequential criterion for continuity proves the required joint continuity throughout the positive quadrant.

For comparison, the change of variables $u=\alpha x^{\beta }$ identifies the value with a gamma function:The domination argument establishes continuity directly, without needing a separate theorem about parameter dependence of the gamma function.

For each nonnegative integer $k$, the change of variables $x=e^{-t}$ givesThe gamma function identity $\Gamma (k+1)=k!$ follows by repeated integration by parts. Since $\log x<0$ on $(0,1)$, the sign of the original integrand is $(-1{)}^k$, soFor $k=0$, the integral is one, consistent with the formula.

The exponential series gives $x^x=e^{x\log x}={\sum }_{k=0}^{\infty }(x\log x{)}^k/k!$. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test proves uniform absolute convergence, using the summable bounds $e^{-k}/k!$. Therefore one can integrate the series term by term over this finite interval. Using part i,This is the alternating identity in sophomore's dream. The same estimate also proves absolute convergence of the integrated series; no cancellation is needed to justify the interchange.

Near zero, $\sin x/x\to 1$, so $|\sin x|/x^{\alpha }$ is comparable to $x^{1-\alpha }$. The power integral test shows that the integral near zero is finite exactly when $\alpha <2$.

For the tail, $|\sin x|\le 1$ proves integrability when $\alpha >1$. To prove necessity, take the disjoint intervalsOn each, $\sin x\ge 1/2$, and $x\le 2\pi (k+1)$. Therefore ${\int }_{I_k}|\sin x|x^{-\alpha }dx\ge c_{\alpha }(k+1{)}^{-\alpha }$ with $c_{\alpha }>0$. The p-series diverges when $\alpha \le 1$, proving that the absolute tail integral diverges in that range.

Combining the two conditions,For $0<\alpha \le 1$, the tail improper integral converges conditionally by the Dirichlet test for improper integrals, but that does not give Lebesgue integrability.

The absolute value is bounded by $e^{-ax}$, whose integral is $1/a$. Thus the function is Lebesgue integrable. The complex integral of an exponential function isTaking the real part, justified by absolute integrability, gives

Again $|e^{-ax}\sin (bx)|\le e^{-ax}$ proves Lebesgue integrability. Taking the imaginary part of the absolutely convergent complex integral in part i gives

The function is not Lebesgue integrable. Fix $0<\eta <\pi /2$ and restrict to $x\ge 1$, $0<y<\eta$. There $\cos y\ge \cos \eta >0$. Tonelli's theorem for the nonnegative absolute value gives the lower boundThe last integral diverges. Indeed, on $[k\pi ,k\pi +\pi /6]$, $|\cos 2x|\ge 1/2$; each such interval contributes at least a positive constant times $1/(k+1)$. The harmonic series diverges. Hence the integral of $|g|$ is infinite, regardless of possible signed cancellations. One cannot infer Lebesgue integrability by formally applying Fubini's theorem to the oscillatory integrand.

The function is Lebesgue integrable. The inequalities $|\sin u|\le \min \{u,1\}$ for $u>0$ give bounds that handle the two ends of the $x$-axis differently. For $0<x<1$,For $x\ge 1$,Tonelli's theorem therefore provesThe factors of sine suppress the near-axis contributions that caused divergence in part i.

Set $C_0=[0,1]$. Inductively, replace each component interval of $C_n$ by its two closed end intervals, each of length ${\ell }_{n+1}:=q_{n+1}/2^{n+1}$. The current component length is ${\ell }_n=q_n/2^n$. Because $q_{n+1}<q_n$, the removed central gap has positive lengthThis is a generalized Cantor set construction. The three required properties are checked in parts i, ii and iii.

The sets are closed, measurable, decreasing, and contained in $[0,1]$, which has finite Lebesgue measure. By continuity from above of a measure,This limit exists because $(q_n)$ is a bounded decreasing monotone sequence. To exclude a nondegenerate interval $J\subseteq C$, note that its connectedness forces it into a single component interval of every $C_n$: crossing two components would cross a removed gap. Its length would therefore be at most $2^{-n}$ for every $n$, which is impossible for a positive-length interval. Thus $C$ contains no nondegenerate interval, even when its Lebesgue measure is positive.

At stage zero there is one closed interval. At every later stage each component is replaced by two closed end intervals separated by a positive open gap. Induction therefore gives exactly $2^n$ pairwise disjoint closed intervals at stage $n$. Their common length issince the decreasing sequence starts at $q_0=1$.

The $2^n$ component intervals are pairwise disjoint and each has length $q_n/2^n$. Finite additivity of Lebesgue measure givesTheir endpoints do not cause overlaps, because the central gaps have strictly positive length.

Each interval retained at stage $n+1$ lies inside the stage-$n$ interval from which it was cut. Taking the union over all parent components proves $\boxed{C_{n+1}\subseteq C_n}$. Consequently the intersection defining $C$ is a nested intersection of nonempty compact sets.

The points belonging to infinitely many $E_n$ form their limit superior of sets:For every $N$, countable subadditivity givesThus $\boxed{m(G)=0}$. Outside this null set, each point belongs to only finitely many $E_n$, so its indicator functions are eventually zero. Therefore $\boxed{{\chi }_{E_n}(x)\to 0\text{ almost everywhere}}$. This is the summable-measure argument behind the first Borel-Cantelli lemma; no independence assumption is needed.

By part i, ${\chi }_{E_n}|g|\to 0$ almost everywhere, while $0\le {\chi }_{E_n}|g|\le |g|$ and $g$ is Lebesgue integrable. The dominated convergence theorem givesTo deduce absolute continuity of the Lebesgue integral, suppose the stated uniform conclusion failed for some ${\epsilon }_0>0$. For each $n$ one could choose a measurable set $A_n$ with $m(A_n)<2^{-n}$ but ${\int }_{A_n}|g|\ge {\epsilon }_0$. Applying the result just proved to this sequence contradicts the lower bound. Hencefor every measurable set $A$.

The a.e. limit $f$ is nonnegative. For a fixed measurable set $A$, Fatou's lemma gives ${\int }_{A}f\le {\mathrm{liminf}}_n{\int }_A{f}_n$. Apply the same lemma to $A^c$ and use convergence of the total integrals:The lower and upper bounds prove $\boxed{{\int }_A{f}_n\to {\int }_{A}f}$.

For the asserted L1 convergence, choose an integer $k$ so large that ${\int }_{|x|>k}f<\epsilon$, possible by Lebesgue integrability. The first result applied to this fixed tail set makes ${\int }_{|x|>k}{f}_n<2\epsilon$ eventually. The additional local hypothesis makes ${\int }_{-k}^k|f_n-f|<\epsilon$ eventually. ThereforeThus $\boxed{\Vert f_n-f{\Vert }_{L^1(\mathbb{R})}\to 0}$.

In fact the extra local hypothesis is unnecessary here. Scheffé's lemma follows directly from $\min \{f_n,f\}\to f$ almost everywhere and domination by $f$:This stronger argument uses precisely the already assumed nonnegativity, a.e. convergence, and convergence of total integrals.