Past exam of the integration course of the University of Oxford

Source: Oxford's released examination PDF, pages 85–87, paper dated 13 June 2023. The numbering below follows the paper; only original solutions are given. In 2(b)(i), the printed integral omits $dx$. In 2(b)(ii), the printed differential equation has a sign error: the correct equation is $F^{\prime }(t)=-2F(t)$, as derived below.

Writing $\mathcal{L}$ for the collection of Lebesgue measurable sets, saying that $\mathcal{L}$ is a sigma-algebra means that $\mathbb{R}\in \mathcal{L}$, that $\mathbb{R}\setminus A\in \mathcal{L}$ whenever $A\in \mathcal{L}$, and that ${\bigcup }_{n=1}^{\infty }{A}_n\in \mathcal{L}$ whenever every $A_n\in \mathcal{L}$. These properties also imply closure under countable intersections and differences.

A real-valued function $f$ is a Lebesgue measurable function precisely when the inverse image of every Borel set is in $\mathcal{L}$. Equivalently, it suffices to test the half-lines:The equivalence follows because these half-lines generate the Borel sigma-algebra, and inverse images preserve complements and countable unions.

The density of the rational numbers gives a countable way to separate two unequal values:Indeed, $f(x)<g(x)$ holds exactly when some rational number lies strictly between them. Every set inside the union is a Lebesgue measurable set, by the definition of a measurable function and the closure properties of a sigma-algebra. Since $\mathbb{Q}$ is countable, the union is measurable.

First suppose $m(E)<\infty$. The interval-cover definition of Lebesgue outer measure, together with $m^{\ast }(E)=m(E)$, provides bounded open intervals $I_j$ covering $E$ such that ${\sum }_j|I_j|<m(E)+\epsilon$. If the definition uses intervals that are not open, enlarge each by an amount whose total is smaller than the available error. For the open set $U={\bigcup }_j{I}_j$, countable subadditivity and additivity of Lebesgue measure giveThis subtraction is valid because both measures are finite.

For arbitrary $E$, including $m(E)=\infty$, set $E_n=E\cap [n,n+1)$ for $n\in \mathbb{Z}$, and choose positive weights $w_n=2^{-|n|}/3$, whose sum is one. Each $E_n$ has finite Lebesgue measure, so the preceding argument gives an open set $U_n\supseteq E_n$ with $m(U_n\setminus E_n)<\epsilon w_n$. Then $U={\bigcup }_{n\in \mathbb{Z}}{U}_n$ is open, contains $E$, andThus the required form of regularity of Lebesgue measure does not require $E$ itself to have finite measure.

The exact criterion isNecessity follows already from one approximation with error less than one: a compact set $K\subseteq \mathbb{R}$ is bounded, hence has finite Lebesgue measure, so $m(E)=m(K)+m(E\setminus K)<\infty$.

Conversely, suppose $m(E)<\infty$. By continuity from above of a measure, the decreasing sets $E\setminus [-N,N]$ have measures tending to zero. Choose $N\ge 1$ so that $m(E\setminus [-N,N])<\epsilon /2$. Put $B=[-N,N]\setminus E$. By part (b)(i), there is an open set $V\supseteq B$ with $m(V\setminus B)<\epsilon /2$. DefineIt is closed and bounded, hence a compact set by the Heine-Borel theorem, and $K\subseteq E$. The discarded portion satisfiesThe second term is at most $m(V\setminus B)$. This proves the claimed inner regularity of Lebesgue measure with an arbitrarily small absolute error.

False. Take $E=\{0\}$ and any non-measurable set $A\subseteq \mathbb{R}$. The vertical line $\{0\}\times \mathbb{R}$ is a planar Lebesgue null set: each segment $\{0\}\times [-N,N]$ can be enclosed in rectangles of arbitrarily small area, and the line is a countable union of these segments. Consequently its subset $E\times A$ is a Lebesgue measurable set of planar measure zero, by Completeness of Lebesgue measure.

A nonempty factor can therefore have too little Lebesgue measure to transmit the other factor's failure of measurability. Its exceptional vertical section at $x=0$ is consistent with Lebesgue measurable sections being guaranteed only almost everywhere.

True. Suppose instead that $B=A_1\times A_2$ were a Lebesgue measurable set in the plane. The section consequence of Fubini's theorem says that there is a Lebesgue null set $N\subseteq \mathbb{R}$ such thatis a Lebesgue measurable set for every $x\notin N$. This version applies to completed Lebesgue measure; it can be obtained on bounded rectangles and then extended by a countable exhaustion, so no finite-measure assumption on $B$ is needed.

For every $x\in A_1$ the section is exactly $A_2$, which is not measurable. Thus $A_1\subseteq N$. By Completeness of Lebesgue measure, every subset of $N$ is measurable, contradicting the hypothesis on $A_1$. Therefore

This requires an argument using Lebesgue outer measure, since $E$ need not be a Lebesgue measurable set. Some bounded portion $F=E\cap [-N,N]$ has $a=m^{\ast }(F)>0$: otherwise countable subadditivity would give $m^{\ast }(E)=0$. Also $a\le 2N<\infty$.

Suppose, for a contradiction, that every nonempty bounded open interval $I$ satisfies $m^{\ast }(E\cap I)\le \lambda |I|$. Because $a/\lambda >a$, the definition of Lebesgue outer measure gives a countable cover $F\subseteq {\bigcup }_j{I}_j$ by nonempty bounded open intervals withUsing countable subadditivity again,a contradiction. Hence at least one interval in such a cover satisfiesThis high-density interval for outer measure applies even to nonmeasurable sets; the argument does not require the Lebesgue density theorem.

The function is continuous on $(0,\infty )$, hence a measurable function. On $(0,1)$ its absolute value is at most one, while on $[1,\infty )$ the elementary estimate $|\sin u|\le |u|$ gives $|\sin (1/x^3)|\le x^{-3}$. By comparison for the Lebesgue integral,It is therefore a Lebesgue integrable function. The oscillations near zero do not obstruct absolute integrability because their amplitude stays bounded on an interval of finite measure.

Again the function is continuous on its domain and therefore a measurable function. For $0<x<1$, $|(\log x)e^{-x}|\le -\log x$, andThis standard improper integral follows from the antiderivative $x-x\log x$ and $x\log x\to 0$ at zero. For $x\ge 1$, $0\le \log x\le x$, and integration by parts gives ${\int }_1^{\infty }x{e}^{-x}dx=2/e$. ThusThe logarithmic singularity is integrable and the exponential factor controls the tail, so this too is a Lebesgue integrable function.

Let $J={\int }_0^{\infty }{e}^{-x^2}dx$. It is positive and finite: on $(0,1)$ the integrand is at most one, and on $[1,\infty )$ it is at most $e^{-x}$. Since the planar integrand is nonnegative and measurable, Tonelli's theorem givesThe change of variables formula in polar coordinates, with $x=r\cos \theta$, $y=r\sin \theta$ and absolute Jacobian determinant $r$, yieldsOne can apply the formula first on $\delta <r<R$, $0<\theta <\pi /2$, where the coordinate map is a diffeomorphism, and then exhaust the quadrant using the monotone convergence theorem. Taking the positive square root proves the half-line Gaussian integral:Here $dx$ is understood in the one-dimensional integral; it is missing from the printed question.

Put $h(t,x)=e^{-x^2-t^2/x^2}$. For every $t\ge 0$, $0<h(t,x)\le e^{-x^2}$, and the upper bound is a Lebesgue integrable function by part (b)(i). Since $h(t,x)\to e^{-x^2}$ for each $x>0$ as $t\downarrow 0$, the dominated convergence theorem gives right continuity at zero andTo apply dominated differentiation under the integral sign, fix $t_0>0$ and restrict $t$ to $[t_0/2,3t_0/2]$. ThenFor $x\ge 1$, $D(x)\le 3t_0{e}^{-x^2}$. For $0<x<1$, writing $a=t_0^2/4>0$, the substitution rule $u=1/x$ givesThus $D$ is integrable. The mean value theorem bounds nearby difference quotients by $D$, so the dominated convergence theorem justifies the derivative at $t_0$. As $t_0$ was arbitrary,Now make the change of variables $u=t/x$. The reversal of limits cancels the negative differential and givesConsequently $F^{\prime }(t)=-2F(t)$. This also follows in sign from the fact that the positive integrand decreases strictly as $t$ increases: the printed equation with a positive sign cannot hold. Solving the linear differential equation, or differentiating $e^{2t}F(t)$, and using right continuity at zero givesThis is a half-line instance of the Gaussian integral with an inverse-square term.

Minkowski inequality states that on a measure space $(X,\Sigma ,\mu )$, for $1\le p\le \infty$ and measurable real or complex $f,g$ of finite Lp norm,For $p=1$, integrate the scalar triangle inequality $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the same result by taking the essential supremum.

For $1<p<\infty$, first ensure that the left side is finite. The convex function $s\mapsto s^p$ on $[0,\infty )$ givesso $f+g$ has finite Lp norm. Set $q=p/(p-1)$ and $H=\Vert f+g{\Vert }_p$. Since $(p-1)q=p$, Holder inequality givesIf $H=0$ the result is immediate; otherwise divide by $H^{p-1}$. This proves the triangle inequality for every claimed exponent.

Use Lebesgue measure restricted to the Lebesgue measurable set $E$. For $1\le p<\infty$, start with the measurable real-valued functions $f$ on $E$ satisfying ${\int }_E|f{|}^{p}dm<\infty$. Define an equivalence relation by $f\sim g$ when $f=g$ almost everywhere, and let $L^p(E)$ consist of their equivalence classes. Addition and scalar multiplication are defined using representatives. The Lp norm isFor $p=\infty$, use measurable functions bounded almost everywhere, apply the same identification, and define the L-infinity norm byThese expressions are independent of the representative. For finite $p$, zero norm is equivalent to $f=0$ almost everywhere: each set $\{|f|>1/n\}$ must have measure zero, and their union is $\{f\ne 0\}$. The same implication follows from the essential supremum for $p=\infty$. Absolute homogeneity is immediate; Minkowski inequality proves the triangle inequality and closure under addition. Thus these are normed vector spaces. Using complex-valued representatives gives their complex versions by the same construction.

Let $(f_n)$ be a Cauchy sequence in the L2 space $L^2(\mathbb{R})$, choosing measurable representatives. There is a subsequence $(f_{n_k})$ withWrite $d_k=f_{n_{k+1}}-f_{n_k}$ and $S_N={\sum }_{k=1}^N|d_k|$. Repeated Minkowski inequality gives $\Vert S_N{\Vert }_2\le {\sum }_{k=1}^N{2}^{-k}\le 1$. As $S_N^2$ increases pointwise, the monotone convergence theorem givesHence $S={\sum }_k|d_k|$ is finite almost everywhere and represents an L2 function. Definewhere the series converges, and set $f=0$ on the exceptional Lebesgue null set. It is a measurable function, and $|f|\le |f_{n_1}|+S$ almost everywhere, so Minkowski inequality gives $f\in L^2(\mathbb{R})$.

For fixed $k$, Fatou's lemma applied to squares of finite tails and then Minkowski inequality giveThus the subsequence converges to $f$ in norm. Given $\epsilon >0$, choose a sufficiently late $n_k$ with $\Vert f-f_{n_k}{\Vert }_2<\epsilon /2$ and with $\Vert f_n-f_{n_k}{\Vert }_2<\epsilon /2$ for every sufficiently large $n$, using the original Cauchy sequence property. The triangle inequality then gives $\Vert f_n-f{\Vert }_2<\epsilon$. ThereforeThis proves the needed instance of completeness of Lp spaces without assuming the result being proved.

False. A typewriter sequence sweeps successively through finer dyadic partitions. For $k\ge 0$ and $j=0,\dots ,2^k-1$, define the indicator functionIts L2 norm satisfiesAs $n$ tends to infinity so does its block index $k$, proving norm convergence to zero. However, every $x\in (0,1)$ belongs to exactly one of the half-open intervals in each block, so $f_n(x)=1$ infinitely often. For $k\ge 1$ there is also an interval in each block that does not contain $x$, so $f_n(x)=0$ infinitely often. Thus there is no pointwise convergence at any $x\in (0,1)$, let alone convergence almost everywhere.

True. Replace the dominating representative by $|g|$ if necessary, and take the countable union of the exceptional Lebesgue null sets so that every bound holds off a single null set. Since the interval has Lebesgue measure one, Cauchy-Schwarz inequality givesThus $g^2$ is a Lebesgue integrable function. The squared functions satisfy $|f_n{|}^2\to 0$ pointwise and $|f_n{|}^2\le |g{|}^2$ almost everywhere. The dominated convergence theorem therefore yieldsThis uses the Lp inclusion on a finite measure space to turn the assumed fourth-power bound into the second-power domination actually needed.

Let $M=\Vert f{\Vert }_{\infty }$, the essential supremum of $|f|$. If $M=0$, $f=0$ almost everywhere and every Lp norm is zero. Suppose $M>0$. Since $|f|\le M$ almost everywhere and $m((0,1))=1$,For any $a$ with $0<a<M$, the measurable set $A_a=\{x\in (0,1):|f(x)|>a\}$ has positive Lebesgue measure; otherwise $a$ would be an essential upper bound. HenceFor fixed $a$, $m(A_a{)}^{1/p}\to 1$. Thus ${\mathrm{liminf}}_{p\to \infty }\Vert f{\Vert }_p\ge a$. Letting $a\uparrow M$ and combining with the upper bound provesThe positive-measure superlevel sets explain why the Lp norms converge to the supremum norm even when the essential maximum is never attained.

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 $A$.