Real and harmonic analysis course of the University of Oxford (C4.2a)

The restriction of a measure to a measurable subset is $(\mu {|}_M)(A)=\mu (A\cap M)$ for every measurable $A$. Absolute continuity of measures, written $\mu \ll \kappa$, means $\kappa (A)=0⟹\mu (A)=0$. The measures are mutually singular when a measurable $S$ satisfies $\kappa (S)=0$ and $\mu (X\setminus S)=0$. Thus singularity means concentration on a null set of the other measure; it does not assert that every measurable set is null for one of them.

Suppose $\mu =\alpha +\sigma =\beta +\tau$ are two such decompositions. Choose measurable supports $S,T$ of the singular parts with $\kappa (S)=\kappa (T)=0$, and put $N=S\cup T$. Absolute continuity makes both $\alpha$ and $\beta$ vanish on $N$. On $X\setminus N$ the two singular parts vanish, so for every measurable $A$,Subtracting from the finite measure $\mu$ gives $\sigma =\tau$. This proves uniqueness of the Lebesgue decomposition without subtracting infinite masses.

For the specified $M$, the restriction to $M$ is singular because $\kappa (M)=0$, and the complementary restriction is absolutely continuous by assumption. Their sum is $\mu$. Uniqueness therefore gives

For the finite nonnegative measures there exist a unique nonnegative integrable density $h\in L^1(\kappa )$, up to $\kappa$-almost-everywhere equality, and a unique nonnegative measure ${\mu }^s\perp \kappa$, such thatThis combines the Lebesgue decomposition theorem with the Radon-Nikodym theorem. In particular ${\mu }^a=h\kappa$ and $h=d{\mu }^a/d\kappa$; if $\mu \ll \kappa$ the singular term is zero.

Write $J(A)$ for the partition supremum. Empty cells are permitted, so $J\ge {\mu }_i$ for each $i$, and $J(A)\le {\sum }_i{\mu }_i(A)$ is finite. Also $J(\varnothing )=0$.

Let $A={⨆}_{j\ge 1}{B}_j$. Any measurable partition $A={⨆}_i{A}_i$ yieldsTaking its supremum gives one inequality. Conversely, for any $\epsilon >0$ choose for each $j$ a partition $B_j={⨆}_i{B}_{ji}$ whose sum exceeds $J(B_j)-\epsilon 2^{-j}$. The sets $A_i={⨆}_j{B}_{ji}$ partition $A$, and countable additivity of each ${\mu }_i$ gives $J(A)\ge {\sum }_{j}J(B_j)-\epsilon$. Let $\epsilon \downarrow 0$. Thus $J$ is a finite measure.

If another measure $\nu$ dominates every ${\mu }_i$, then for every such partition ${\sum }_i{\mu }_i(A_i)\le {\sum }_i\nu (A_i)=\nu (A)$. Consequently $J\le \nu$. Hence $J={\bigvee }_i{\mu }_i$ is the least dominating measure, the finite supremum of positive measures. It generally differs from the pointwise maximum of the numbers ${\mu }_i(A)$: the maximizing index may differ between pieces of $A$.

Choose measurable sets $S_i$ carrying ${\mu }_i^s$ with $\kappa (S_i)=0$, and put $S={\bigcup }_i{S}_i$. Every absolutely continuous component vanishes on $S$, and every singular component vanishes on its complement. The given restriction identity therefore givesThe first is singular with respect to $\kappa$. The second is absolutely continuous because it is bounded above by ${\sum }_i{\mu }_i^a$. They sum to $\mu$, so uniqueness identifies them as its Lebesgue components.

Let $h_i=d{\mu }_i^a/d\kappa$ and $h={\max }_i{h}_i$, using measurable representatives. Each partition sum is at most ${\int }_{A}hd\kappa$. For the reverse inequality, partition $A$ measurably by assigning a point to the smallest index attaining this finite maximum. On its assigned cell $h_i=h$, and the partition sum is exactly ${\int }_{A}hd\kappa$. ThusThe maximum of finitely many integrable nonnegative densities is integrable, since it is at most their sum.

The Lebesgue outer measure iswhere $|Q|$ is the product of its side lengths. The infimum ranges over countable covers; allowing other standard endpoint conventions gives the same outer measure by arbitrarily small enlargement. Lebesgue measure is its restriction to Lebesgue measurable sets.

Enlarged unit cubes give ${\mathcal{L}}^{\ast }([0,1{]}^n)\le 1$. For any countable open-box cover of this compact cube choose a finite subcover. A finite grid formed from its box endpoints and the cube endpoints partitions the cube into rectangular cells. The sum of the volumes of covering boxes is at least the sum of these cell volumes, which is one. Boundary faces have elementary zero volume and do not affect the finite volume comparison. Thus every covering sum is at least one, proving

A set $H$ is Lebesgue measurable when the Caratheodory measurability criterion holds:Subadditivity always gives the inequality with left side at most right side. For $H=\{x_i>t\}$, split each covering box $Q_j$ across the plane $x_i=t$. Its two pieces have total volume $|Q_j|$. The piece with a closed cut face can be enlarged by arbitrarily little into an open box; hence its outer measure is at most its corresponding volume, just as for the other piece. Subadditivity applied separately on the two sides consequently givesFor finite outer measure take a covering sum arbitrarily close to ${\mathcal{L}}^{\ast }(E)$; infinite outer measure follows already from subadditivity. Thus every coordinate half space is measurable.

Carathéodory's theorem says that the measurable sets form a sigma-algebra and the outer measure restricted to it is a complete measure. Coordinate half spaces generate all open boxes and then all open sets, using countably many rational-endpoint boxes. They therefore generate the Borel sigma-algebra. Hence every Borel set is Lebesgue measurable.

A regular outer measure has a measurable hull of equal outer measure for every set: for each $A$ there is measurable $G\supseteq A$ with ${\mathcal{L}}^{\ast }(G)={\mathcal{L}}^{\ast }(A)$. Lebesgue outer measure has even a Borel hull. If its value on $A$ is finite, choose open-box covers of total cost below ${\mathcal{L}}^{\ast }(A)+1/j$, and let $U_j$ be their open unions. Then $G={\bigcap }_j{U}_j$ is Borel andFor infinite outer measure the whole space is a suitable hull. Thus Lebesgue outer measure is regular.

If $A$ is measurable, additivity immediately gives the requested equality. Conversely suppose the equality holds. Choose a measurable hull $G$ of $A$ and put $B=G\cap M$. Then $A\subseteq B\subseteq M$ and $\mathcal{L}(B)={\mathcal{L}}^{\ast }(A)$. Since $B$ is measurable, its splitting of $M\setminus A$ givesThe assumed equality and $\mathcal{L}(M)=\mathcal{L}(B)+\mathcal{L}(M\setminus B)$ imply ${\mathcal{L}}^{\ast }(B\setminus A)=0$. These subtractions are legitimate because $\mathcal{L}(M)<\infty$. Outer-measure-null sets are measurable by the Caratheodory criterion. Hence $A=B\setminus (B\setminus A)$ is measurable, proving the equivalence.

For a complex measure, its total variation measure isIts norm is $\Vert \mu \Vert =|\mu |(\mathbb{T})$. For Radon measures on the compact circle this is also ${\sup }_{\Vert f{\Vert }_{\infty }\le 1,f\in C(\mathbb{T})}|\int fd\mu |$. The stated convergence is weak-star convergence against continuous tests:It is the weak-star topology from the identification of these measures with the dual of $C(\mathbb{T})$, rather than convergence of their values on every Borel set.

Set $C={\sup }_k\Vert {\mu }_k\Vert$. Given $f\in C(\mathbb{T})$, choose $g\in X$ with $\Vert f-g{\Vert }_{\infty }<\epsilon$. ThenTake the upper limit in $k$ and then $\epsilon \downarrow 0$. Thus convergence on the dense test family suffices; no assumption that $X$ is a vector space is needed.

For nonempty compact $K$, put $f_j(x)=\max \{0,1-j\mathrm{dist}(x,K)\}$. These continuous functions decrease to $1_K$ and satisfy ${\mu }_k(K)\le \int f_{j}d{\mu }_k$. Consequently ${\mathrm{limsup}}_k{\mu }_k(K)\le \int f_{j}d\mu$ for every $j$. Dominated convergence for the finite positive measure $\mu$ yieldsThe empty-set case is immediate. Positivity is essential for the comparison with these cutoff functions.

Write $m(dx)=dx/(2\pi )$ for the normalized Haar measure. The identity ${\sin }^2(kx)=\frac{1}{2}-\frac{1}{2}\cos (2kx)$ gives ${\mu }_k=\frac{1}{2}m-\frac{1}{2}\cos (2kx)m$. For any fixed trigonometric polynomial, orthogonality of complex exponentials makes the oscillatory integral zero once $2k$ exceeds its largest frequency. The norms satisfy $\Vert {\mu }_k\Vert ={\mu }_k(\mathbb{T})=1/2$, so the dense-test criterion applies. ThereforeIn particular the limit has mass one half; treating these measures as probabilities would introduce a factor-of-two error.

The Fourier coefficient of a measure on the circle and convolution of measures on the circle use the angular group law:Addition is modulo $2\pi$. This is the pushforward of the product complex measure under addition; equivalently its integral against a continuous $f$ is $\iint f(x+y)d\mu (x)d\nu (y)$. The integrands are bounded and the variation measures finite, so Fubini applies. In particular,Thus $\widehat{\mu \ast \nu }(n)=\hat{\mu }(n)\hat{\nu }(n)$. These measure coefficients use no additional Haar normalization factor: a probability has zeroth coefficient one.

Read the coefficient inside the limit as $\widehat{{\mu }_k}(n)$; the printed omission of its index would otherwise make that condition independent of the sequence. Weak-star convergence immediately implies convergence of every coefficient by testing $e^{-inx}$. Conversely coefficient convergence gives convergence against every trigonometric polynomial. Since $\Vert {\mu }_k\Vert =1$, the dense-test argument yields weak-star convergence to $\mu$.

The limit is necessarily a probability. Testing the constant function gives $\mu (\mathbb{T})=1$. For every nonnegative continuous $f$, $\int fd\mu ={\lim }_k\int fd{\mu }_k\ge 0$. A Radon measure whose integrals against all such functions are nonnegative is positive, by the positive case of the Riesz representation theorem (or continuous approximation of Borel indicators). Thus $\mu$ is nonnegative and has total mass one.

Convolution preserves probability measures: its defining pushforward is nonnegative and its total mass is the product of the two unit masses. For every fixed integer $n$ the previous coefficient identity givesThe probability version of the Fourier coefficient criterion therefore provesThe fixed compact underlying group and the uniform probability masses supply precisely the bounds needed in the dense-test argument.

The centered Hardy-Littlewood maximal function isThe Hardy-Littlewood maximal inequality gives, for example,This valid constant follows from the disjoint-interval covering argument: a compact subset of the superlevel set has a finite cover by intervals of average above $t$; select disjoint intervals whose triples cover it, bound their total length by $\Vert f{\Vert }_1/t$, and use inner regularity. An optimal constant is not required.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.

For a set $M$ of finite measure apply the preceding result to $f=1_M$. Since its local average minus its point value is bounded in absolute value by its mean absolute oscillation, it givesFor a general measurable $M$, apply the same argument to $1_{M\cap [-j,j]}\in L^1(\mathbb{R})$. At every interior point of $(-j,j)$, sufficiently short averages agree with those of $1_M$. The countable union of these interiors covers the line, and a countable union of null exceptional sets remains null. This proves the Lebesgue density theorem without a finite-measure assumption on $M$. The source's $\chi$ is the indicator of $M$.

Every $x$ in the maximal superlevel set has a first index $k(x)$ at which the average on its unique cell $M_k(x)\in {\mathcal{P}}_k$ exceeds $t$. Refinement implies that all earlier cells containing $x$ contain this cell, and their averages are at most $t$. Thus $M_{k(x)}(x)\in {\mathcal{S}}_t$ and the selected cells cover the superlevel set.

Two cells at different levels are either disjoint or nested, because of refinement and the partition property. If selected cells of levels $k<l$ intersect, the level-$l$ one lies in the earlier cell whose average is above $t$, contradicting its first-crossing property. Cells at the same level are disjoint unless identical. Hence the selected cells form a countable disjoint partition of the superlevel set. A cell repeated unchanged at successive levels is counted once, at its first crossing; equal earlier ancestors are included in the comparison condition.

The first-crossing maximal inequality for refining partitions follows from countable disjointness and the strict average bound:Thus this partition maximal operator has weak-type constant one.

The conclusion of differentiation along refining metric partitions isThe crucial geometric property is shrinking diameter, not merely refinement: since $x\in M_k(x)$ and $\mathrm{diam}{M}_k(x)<1/k$, continuity makes the corresponding oscillation tend to zero for every continuous test function. Continuous $L^1$ approximation and the maximal inequality then transfer this to $f$, exactly as before: the exceptional-set bound is $2\Vert f-g{\Vert }_1/\epsilon$ for an oscillation threshold $2\epsilon$. Countability makes the maximal function measurable, and positive cell measure makes every average defined. Refinement supplied the disjoint first-crossing argument; shrinking diameter supplies differentiation.

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.