Past exam of the real and harmonic analysis course of the University of Oxford / 2007 (C4.2a, DMAC 2A79, DMPC 2A79)

The absolute continuity of measures condition isThe Radon-Nikodym theorem says that if both positive measures are sigma-finite measures, then this condition is equivalent to the existence of a nonnegative measurable function $h$, unique up to equality $\kappa$-almost everywhere, such thatOne writes $h=d\nu /d\kappa$, the Radon-Nikodym derivative. Under these hypotheses $h$ is finite $\kappa$-almost everywhere. If $\nu$ is finite, then $h\in L^1(\kappa )$. A finite signed measure absolutely continuous with respect to a sigma-finite positive measure likewise has an integrable signed Radon-Nikodym derivative.

The sigma-finiteness hypothesis on the dominating measure cannot simply be omitted. On the Borel subsets of $[0,1]$, let $\kappa$ be counting measure and $\nu$ Lebesgue measure. Absolute continuity holds because the only $\kappa$-null set is empty. A Radon-Nikodym derivative would satisfy $h(x)=\nu (\{x\})=0$ at every point, contradicting $\nu ([0,1])=1$. Here counting measure is not sigma-finite.

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.

From now on $|\mu |(X)<\infty$, so all signed quantities below are finite. Since $|\mu |(E)\ge |\mu (E)|$, the set functionis nonnegative. Countable additivity follows from that of the two terms, so ${\mu }^{+}$ is a finite positive measure.

Set $s=\sup \{|\mu |(M):{\mu }^{+}(M)=0\}$. This is finite. Choose measurable $M_j$ with ${\mu }^{+}(M_j)=0$ and $|\mu |(M_j)\to s$. Thenis ${\mu }^{+}$-null by countable subadditivity, and monotonicity gives $|\mu |(M_{-})\ge s$. The definition of $s$ gives the reverse inequality. Hence $M_{-}$ attains the maximum, rather than merely approximating it.

Start with $R_0=A$. Having chosen disjoint $A_1,\dots ,A_i$, put $R_i=A\setminus {\bigcup }_{j\le i}{A}_j$ andThe empty set shows $s_i\ge 0$, and finite total variation bounds it above. If $s_i>0$, choose $A_{i+1}\subseteq R_i$ with $\mu (A_{i+1})\ge s_i/2>0$; if $s_i=0$, choose $A_{i+1}=\varnothing$. This is the required recursion, including the initial step $i=0$.

Disjointness and nonnegativity give ${\sum }_i\mu (A_i)\le |\mu |(A)<\infty$, hence $\mu (A_i)\to 0$. Write $R=A\setminus {\bigcup }_i{A}_i$. For every measurable $B\subseteq R$,Thus every measurable subset of $R$ has nonpositive measure. Each finite partition of $R$ consequently has variation sum exactly $-\mu (R)$, and therefore $|\mu |(R)=-\mu (R)$. It follows that ${\mu }^{+}(R)=0$. Finally countable additivity gives

If $S\subseteq M_{-}$, then ${\mu }^{+}(S)=0$, so $\mu (S)=-|\mu |(S)\le 0$. Suppose instead that $S\cap M_{-}=\varnothing$ but $\mu (S)<0$. The preceding construction gives $R\subseteq S$ with ${\mu }^{+}(R)=0$ and $|\mu |(R)=-\mu (R)>0$. Then $M_{-}\cup R$ is ${\mu }^{+}$-null with variation strictly larger than that of $M_{-}$, a contradiction. Hence $\mu (S)\ge 0$ on the complement of $M_{-}$. This is a Hahn decomposition.

On $M_{+}=X\setminus M_{-}$ every partition has nonnegative signed measures, so $|\mu |(E)=\mu (E)={\mu }^{+}(E)$ for $E\subseteq M_{+}$. On $M_{-}$, ${\mu }^{+}$ vanishes. Splitting an arbitrary measurable set between these two sets givesIn particular ${\mu }^{+}\ll |\mu |$, and the exhibited Radon-Nikodym derivative verifies the Radon-Nikodym theorem for this pair directly.

An outer measure on $X$ is a function $\mu :\mathcal{P}(X)\to [0,\infty ]$ satisfying $\mu (\varnothing )=0$, monotonicity, and countable subadditivity:Its measurable sets are those satisfying Caratheodory measurability criterion:The test runs over all subsets $E$, including nonmeasurable ones.

The Caratheodory measurability criterion yields a complete sigma-algebra ${\mathfrak{M}}_{\mu }^{\ast }$, and $\mu$ restricted to it is a countably additive measure. Completeness means that every subset of a measurable null set is measurable.

For the requested algebra proof, $X$ and $\varnothing$ plainly satisfy the criterion. Replacing $A$ by $X\setminus A$ merely exchanges the two terms, proving closure under complements. If $A,B$ are measurable, apply their splitting identities successively to any $E$:Subadditivity applied to the first two pieces impliesThe reverse inequality is outer subadditivity. Hence $A\cup B$ is measurable, proving closure under finite unions and therefore the algebra of sets property. The argument never subtracts infinite values.

A metric outer measure on a metric space $(X,d)$ is an outer measure satisfyingHere $d(A,B)=\inf \{d(a,b):a\in A,b\in B\}$. Every Borel set is measurable for a metric outer measure. Consequently its measurable sigma-algebra contains the Borel sigma-algebra; it may be strictly larger.

For $E\subseteq {\mathbb{R}}^n$, define the Lebesgue outer measure bywhere the boxes are bounded and $\mathrm{vol}(Q_j)={\prod }_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. Monotonicity follows by inclusion of admissible covers. Taking covers within $\epsilon 2^{-j}$ of the outer measures of a sequence of sets proves countable subadditivity; when the sum is infinite the inequality is automatic. Lebesgue measure is the restriction of ${\mathcal{L}}^{\ast }$ to its Caratheodory-measurable sets.

If $d(A,B)=\delta >0$, subdivide each box of any countable cover of $A\cup B$ into finitely many half-open boxes of diameter smaller than $\delta$, preserving its total volume. No small box can meet both $A$ and $B$. The boxes meeting $A$ cover $A$, and the disjoint family meeting $B$ covers $B$, so the original cover costs at least ${\mathcal{L}}^{\ast }(A)+{\mathcal{L}}^{\ast }(B)$. Taking the infimum and then using subadditivity givesThus the outer measure underlying Lebesgue measure is a metric outer measure, and in particular every Borel set is Lebesgue measurable.

A regular outer measure has, for every $E\subseteq X$, a measurable hull $H\supseteq E$ with $\mu (H)=\mu (E)$.

Suppose $\mathcal{C}$ is an algebra containing ${\mathfrak{M}}_{\mu }^{\ast }$ and $\mu$ is finitely additive on $\mathcal{C}$. Fix $A\in \mathcal{C}$ and an arbitrary $E$ of finite outer measure. Take a measurable hull $H$ of $E$. Both $H\cap A$ and $H\setminus A$ belong to $\mathcal{C}$, soOuter subadditivity gives equality. If $\mu (E)=\infty$, outer subadditivity already forces the sum on the right to be infinite, again giving equality. Therefore $A$ satisfies the Caratheodory criterion. We have proved $\mathcal{C}\subseteq {\mathfrak{M}}_{\mu }^{\ast }$, soThis proves the maximal additivity algebra of a regular outer measure without any sigma-finiteness assumption.

The Riesz-Markov-Kakutani representation theorem identifies the continuous dual of $C(\mathbb{T})$ with the finite complex Radon measures on $\mathbb{T}$: every bounded complex-linear functional $L$ has a unique representationThe right side is the total variation norm of the measure. On this compact metric space Radon measures are regular Borel measures. A functional is positive precisely when its representing measure is positive; then $\Vert L\Vert =L(1)=\mu (\mathbb{T})$.

Let $m$ be normalized arclength on the unit circle, so in angular coordinates $dm=dt/(2\pi )$. It is a finite Radon measure. The given identity says that $\mu$ and $m$ represent the same continuous linear functional on $C(\mathbb{T})$. Uniqueness in the Riesz-Markov-Kakutani representation theorem givesThe printed phrase “Borel sets in $E$” is interpreted as Borel subsets $E$ of the circle.

Write $A_N(f)=(N+1{)}^{-1}{\sum }_{n=0}^{N}f(e^{in\xi })$. For a monomial $z^k$, $k\in \mathbb{Z}$, the geometric series givesThus every such average converges, and by linearity so does the average of every trigonometric polynomial. Also $|A_N(f)|\le \Vert f{\Vert }_{\infty }$ uniformly in $N$.

Given $\epsilon >0$, choose a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. ThenThe last term tends to zero as $N,M\to \infty$. The scalar sequence is Cauchy and therefore the limit exists for every continuous $f$ and every real $\xi$.

Define $L_{\xi }(f)={\lim }_N{A}_N(f)$. Limits preserve linearity and positivity, and $|L_{\xi }(f)|\le \Vert f{\Vert }_{\infty }$, $L_{\xi }(1)=1$. The Riesz-Markov-Kakutani representation theorem supplies a unique Radon probability measure ${\mu }^{\xi }$. By its definition,which is exactly weak-star convergence in $C(\mathbb{T}{)}^{\prime }$.

If $\xi /(2\pi )=p/q$ in lowest terms, the orbit has exactly $q$ points. Averaging a periodic sequence givesThis includes $q=1$. If $\xi /\pi$ is irrational, then $e^{ik\xi }\ne 1$ for every nonzero integer $k$. The preceding calculation gives $L_{\xi }(z^k)=0$ for $k\ne 0$ and $L_{\xi }(1)=1$, exactly the integrals against normalized Haar measure $m$. Density of trigonometric polynomials implies equality on all continuous functions, and uniqueness givesThese are the periodic-orbit and irrational-rotation cases of equidistribution.

For each $n\ge 0$, choose an open arc $I_n$ about $e^{in}$ of arclength $\pi 2^{-n-2}$, and put $G^1={\bigcup }_{n\ge 0}{I}_n$. It contains every point used by every empirical measure, whereasThus ${\mu }_N^1(G^1)=1$ for all $N$. Let $K=\mathbb{T}\setminus G^1$. This is closed and ${\mu }_N^1(K)=0$ for all $N$. But $1/\pi$ is irrational, so the weak-star limit is normalized arclength andTherefore the proposed implication is false. Weak-star convergence tests continuous functions and does not assert convergence of the indicator function of an arbitrary closed set.

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

A summability kernel on the circle is a sequence of continuous periodic functions withThe frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For $f$ in a homogeneous Banach space,Given $\epsilon >0$, translation continuity bounds the difference inside a sufficiently small neighborhood by $\epsilon$; its integral is at most $C\epsilon$. Outside, use $2\Vert f{\Vert }_B$ and the vanishing tail condition. Hence $f\ast k_n\to f$ in $B$.

With the indexing needed in part (c), the Fejér kernel isIn particular $K_0=1$. The squared-modulus expression proves $K_n\ge 0$, and its constant Fourier coefficient proves $\int K_{n}dm=\Vert K_n{\Vert }_1=1$. The geometric series yieldsThus these are summability kernels. For $T_f:g\mapsto f\ast g$, part (a) gives $\Vert T_f\Vert \le \Vert f{\Vert }_B$. Since $\Vert K_n{\Vert }_1=1$ and $T_f{K}_n\to f$ in $B$, the reverse bound follows:

For $n\ge 1$ put $q_n(t)=2n{K}_{n-1}(t)\sin (nt)$. Using $\sin (nt)=(e^{int}-e^{-int})/(2i)$ and the Fourier coefficients of $K_{n-1}$ givesIndeed, when $1\le k\le n$ only the $e^{int}$ shift contributes, with coefficient $(-in)(1-|k-n|/n)=-ik$. For $-n\le k\le -1$ only the opposite shift contributes and gives $in(1-|k+n|/n)=-ik$; at $k=0$ both coefficients vanish.

If $P={\sum }_{|k|\le n}{a}_k{e}^{ikt}$, convolution multiplies its Fourier coefficients by those of $q_n$. Thus $-P\ast q_n=\sum ik{a}_k{e}^{ikt}=P^{\prime }$, proving the exact identity. Since $K_{n-1}\ge 0$ and has integral one,For $n=0$ the polynomial is constant and the bound is immediate. This elementary Fejér-kernel argument proves the requested estimate; the sharp Bernstein inequality for trigonometric polynomials has constant $n$.