Past exam of the integration course of the University of Oxford / 2015 (A4, A13756W1)

Source release: Oxford examination archive, member FOI-20210708-1 - Data/a13756w1 (4).pdf. The cover dates the examination to 18 June 2015.

Answer availability checked on 28 October 2026: a bounded search of official Oxford sources found no advertisement of complete or selected-part solutions for this exact paper. Oxford's stated policy excludes odd examination years apart from the exceptional release for 2019.^[ref]

A measurable space is a pair $(\Omega ,\mathcal{F})$ where $\mathcal{F}$ is a sigma-algebra of subsets of $\Omega$: it contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. A measure on it is a function $\mu :\mathcal{F}\to [0,\infty ]$ with $\mu (\varnothing )=0$ and countable additivity:whenever the measurable sets $A_n$ are pairwise disjoint. Infinite values are allowed. The triple $(\Omega ,\mathcal{F},\mu )$ is a measure space.

Because $E\in \mathcal{F}$, the family ${\mathcal{F}}_E$ contains $E$. If $A\in {\mathcal{F}}_E$, then $E\setminus A=E\cap (\Omega \setminus A)$ belongs to $\mathcal{F}$ and is a subset of $E$, so it belongs to ${\mathcal{F}}_E$. If each $A_n\in {\mathcal{F}}_E$, their countable union belongs to $\mathcal{F}$ and remains a subset of $E$. Thus ${\mathcal{F}}_E$ is a sigma-algebra on $E$, with relative complements taken in $E$, not in $\Omega$.

Equivalently,the trace sigma-algebra on the measurable set $E$. The measurability of $E$ is what makes this equal to the family in the question.

The function $\nu$ takes values in $[0,\infty ]$ and has $\nu (\varnothing )=0$. For any pairwise disjoint $A_n\in {\mathcal{F}}_E$, countable additivity of $\mu$ givesTherefore $\nu$ is a measure on $(E,{\mathcal{F}}_E)$: it is the restriction of a measure to a measurable subset.

For the increasing family, define the measurable setsThey are pairwise disjoint, with $A_N={\bigcup }_{n=1}^N{D}_n$ and ${\bigcup }_n{A}_n={\bigcup }_n{D}_n$. By countable additivity,The nonnegative series is defined by its increasing partial-sum limit, so this proof also covers an infinite value. This is continuity from below of a measure.

For the decreasing family, put $C_n=B_1\setminus B_n$. These measurable sets increase, andSince $\mu (B_1)<\infty$, finite additivity gives legitimate finite subtractions:Apply the first result to $C_n$ and subtract from $\mu (B_1)$ to obtainThis is continuity from above of a measure. Its finiteness hypothesis matters: for Lebesgue measure and $B_n=[n,\infty )$, every $B_n$ has infinite measure but their intersection is empty. The proof must not subtract $\infty -\infty$.

For $x<y$, finite additivity on the disjoint pieces of the half-line givesBy monotonicity of a measure and the Lebesgue measure of an interval,Exchanging the two points if necessary provesThus the cumulative measure of a finite-measure set is a Lipschitz function with Lipschitz constant at most one, hence is uniformly continuous: given $\epsilon >0$, the choice $\delta =\epsilon$ works simultaneously at every point.

To locate its endpoint limits, the measurable sets $E\cap (-\infty ,N]$ increase to $E$, so continuity from below of a measure gives $h(N)\to m(E)$. The measurable sets $E\cap (-\infty ,-N]$ decrease to the empty set and all have finite measure, so continuity from above of a measure gives $h(-N)\to 0$. Since $h$ is nondecreasing,For $0<c<m(E)$, choose $R>0$ with $h(-R)<c<h(R)$. The intermediate value theorem on $[-R,R]$ gives $t$ with $h(t)=c$. ThenThe endpoint values $c=0,m(E)$ can also be achieved by $A=\varnothing ,E$. This proves divisibility of Lebesgue measure for arbitrary finite-measure Lebesgue measurable sets on the real line, including unbounded ones. The non-atomic measure structure permits prescribed fractions of measure; a measure concentrated at one point would not have this property.

The function is continuous for $x>0$, hence Lebesgue measurable. Near zero, the bound $|\sin x|\le x$ givesThus $f$ is Lebesgue integrable on $(0,1]$ for $q<2$. The convergence is an instance of power singularity integrability.

For the tail, take the pairwise disjoint intervalsOn each $I_k$, $|\sin x|\ge 1/2$, $x\le (k+1)\pi$, and the interval length is $2\pi /3$. ThereforeSumming the nonnegative integrals over these disjoint intervals yieldsby divergence of the p-series. Hence $f$ is not Lebesgue integrable on $[1,\infty )$ for $0<q\le 1$.

The test is absolute integrability: oscillatory sign cancellation does not make the absolute-value integral finite. A conditionally convergent improper integral is a different notion. The sine power kernel integrability criterion separates these endpoint thresholds.

DefineThis is a Lebesgue integrable function: on $(0,1]$ it is bounded by $x^{-1/2}$, whose integral is two, while on $[1,\infty )$ it is bounded by $e^{-x}$, whose integral is $e^{-1}$. Since $|\sin (xy)|\le x|y|$,Thus $h_y$ is Lebesgue integrable for every $y\ge 0$, including $h_0=0$; the same argument actually works for all real $y$.

The mean value theorem for the sine gives $|\sin u-\sin v|\le |u-v|$. Henceand the triangle inequality for an integral givesThus $F$ is a globally Lipschitz function, in particular continuous on $[0,\infty )$, including the endpoint zero.

For the derivative, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the integrandFor every $x>0$, $D_t(x)\to H(x)\cos (xy)$ as $t\to 0$. The same mean value theorem gives the uniform-in-$t$ bound $|D_t(x)|\le H(x)$.

The dominated convergence theorem states that if measurable functions $u_n$ converge almost everywhere to $u$ and $|u_n|\le G$ for a single Lebesgue integrable function $G$, then $u$ is Lebesgue integrable and $\int |u_n-u|\to 0$, hence $\int u_n\to \int u$. Apply it to the difference quotients along every sequence $t_n\to 0$, using the common dominator $H$. The resulting limit is independent of the sequence, soThe right-hand integral is absolutely convergent because its integrand has absolute value at most $H$. This is dominated differentiation under the integral sign, justified by an explicit integrable bound rather than a formal interchange.

For every fixed $x>0$, $\log (x+n)/n\to 0$, so $f_n(x)\to 0$. To exchange the limit with the Lebesgue integral, we need a common integrable bound over the whole half-line.

For $n\ge 1$, $x+n\le n(1+x)$, soTogether with $|\sin x|\le \min (x,1)$, this givesFor $0<x\le 1$, $\log (1+x)\le x\le 1$, hence $G(x)\le 2x^{1-q}$, which is integrable because $q<2$. For $x\ge 1$,Both terms are integrable because $q>1$:The second equality follows from integration by parts, or the substitution $x=e^t$. Consequently $G$ is a Lebesgue integrable function, and every $f_n$ is Lebesgue integrable.

The dominated convergence theorem states that almost everywhere convergence of measurable functions under one integrable absolute-value bound implies L1 convergence, and in particular convergence of their integrals. Here the pointwise limit is zero and the dominator is $G$, soIn fact we obtain the stronger conclusion $\Vert f_n{\Vert }_1\to 0$. A quantitative version follows by retaining the factors $1/n$ in the bound:The preceding endpoint estimates show $A,B<\infty$, so the L1 norm is $O(\log n/n)$ as $n\to \infty$, with $O$ denoting Big O notation. This is logarithmic reweighting under dominated convergence.

Fix $y\ne 0$. The function $x\mapsto f(x,y)$ is continuous on $[-1,1]$, because its denominator is bounded away from zero, and it is an odd function of $x$. Its integral over this symmetric interval is therefore zero. At $y=0$ the entire section is zero, including the specified value at the origin. Thus the inner $x$ integral exists for every $y$ and is identically zero. The same reasoning with $x$ and $y$ exchanged givesTo test Lebesgue integrability in two dimensions, examine the absolute value near the origin. For $0<\epsilon <1$, the annulus $\epsilon <\sqrt{x^2+y^2}<1$ lies inside the square. In polar coordinates,The change of variables formula on this annulus, where the integrand has no singularity, givesThe lower bound tends to infinity as $\epsilon \downarrow 0$. HenceMoreover reflection in either coordinate axis exchanges the positive part and negative part while preserving Lebesgue measure. Both parts therefore have infinite integral, so a signed double Lebesgue integral would involve the undefined expression $\infty -\infty$.

There is no contradiction with Fubini's theorem: its absolute integrability hypothesis is missing. This is an example of finite iterated integrals without joint integrability, with every individual one-dimensional section integrable despite the two-dimensional singularity.

Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.