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A positive-definite function $r$ on the discrete group $\mathbb{Z}$ is the sequence of Fourier coefficients of a unique finite positive Borel measure on the circle, with total mass $r(0)$. Positivity makes the finite sums ${\sum }_{|n|<N}(1-|n|/N)r(n)e^{-2\pi int}$ nonnegative. Their measures converge weakly to the required measure; uniqueness follows from density of trigonometric polynomials in continuous functions.

A Borel probability measure is a Borel measure with total mass one. On a compact metric space it is determined by its integrals against continuous functions. On that compact metric space, the family of Borel probability measures is compact for weak convergence of probability measures.

A positive Borel measure on the unit disc satisfies $\int |f|d\mu \le N\Vert f{\Vert }_1$ for every H1 space on the unit disk function exactly when it is a Carleson measure, with comparable optimal constants. The normalized squared Cauchy kernel tests necessity. For sufficiency, the boundary shadow of every point with $|f(z)|>t$ lies inside a component arc of $\{Nf>t\}$. The geodesic-cap and Carleson-box comparison bounds the measure of each shadow region by its component length, giving $\mu \{|f|>t\}\le C\Vert \mu {\Vert }_{\mathrm{Carl}}m\{Nf>t\}$. The layer cake representation and H1 non-tangential maximal inequality on the disk give the embedding. The broader kernel and covering mechanism is discussed in Terence Tao's harmonic-analysis notes, Exercise 2 after Lemma 4.1; the analytic endpoint here uses the fractional-power argument just given.

A positive Borel measure near a domain boundary is a Carleson measure when its mass in boundary neighborhoods is bounded by a constant times their boundary measure. On the unit disc, this becomes the displayed bound for every Carleson box. The equivalent hyperbolic geodesic cap version follows from the geodesic-cap and Carleson-box comparison. The Carleson embedding theorem on the disk explains why this geometric growth condition controls integration of analytic Hardy spaces.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

If finite Borel measures $\mu$ and $\nu$ on $\mathbb{R}$ satisfyfor every compactly supported continuous function $f$, then $\mu =\nu$. This follows from the uniqueness part of the Riesz representation theorem; alternatively, continuous functions that approximate indicators from above and below recover the measure of every interval and hence every Borel set.

The completion of a measure extends a measure space $(X,\mathcal{F},\mu )$ so that every subset of a measurable null set is measurable. Its completed sigma-algebra isand the extension is $\overline{\mu }(A\cup N)=\mu (A)$. This is well-defined: two such representations differ only inside a measurable null set, so their measurable parts have the same measure. The extension agrees with $\mu$ on $\mathcal{F}$ and is a complete measure.^[ref]

For example, completing the Borel measure given by length on the real line gives Lebesgue measure on the Lebesgue measurable sets, including every subset of every Borel null set. Completion of a measure changes which sets are measurable; it does not identify distinct sets. If $A$ and $B$ differ only on a null set, they can still be different sets in the completed sigma-algebra. Equality modulo null sets is an additional quotient construction.

If a finite complex Borel measure $\mu$ on the circle has Fourier coefficients $\hat{\mu }(n)=0$ for all $n<0$, then it is absolutely continuous with respect to arc length. The reflected Fourier convention gives the equivalent theorem with positive coefficients vanishing. The Poisson integral on the unit disk is analytic because of the one-sided Fourier spectrum. Subharmonic comparison for its square root and an $L^2$ maximal estimate produce an integrable radial maximum. This maximum dominates the weak-star boundary measure, forcing absolute continuity of measures. The same proof on the real line says that a finite measure with Fourier transform of a finite measure supported in $[0,\infty )$ has a Lebesgue density.

A left Haar measure on a locally compact group that is a Hausdorff space is a nonzero regular Borel measure, finite on compact sets, satisfying $\mu (gA)=\mu (A)$ for every Borel set $A$. Such a measure exists and is unique up to a positive scalar. Right Haar measure instead satisfies $\mu (Ag)=\mu (A)$; the two need not coincide. On a compact group, normalization gives total mass one and invariance on both sides. See Tornier, Haar Measures, Definition 2.1 and Theorem 2.2.

For a finite group, normalized Haar measure is uniform counting; for the circle group, it is normalized angle integration on the circle. These examples turn symmetry into a rule for averaging.

For the upper half-space $H^{d+1}={\mathbb{R}}^d\times (0,\infty )$, this is the space of harmonic functions with finite norm $\Vert u{\Vert }_{h^1}={\sup }_{t>0}{\int }_{{\mathbb{R}}^d}|u(x,t)|dx$. It differs from an analytic Hardy space: no holomorphic function condition is imposed, and its boundary object can be a singular finite Borel measure.

For probability measures defined as Borel measures on Polish spaces and a nonnegative sequentially lower semicontinuous cost, the minimum cost over transport plans equals the supremum of $\int ud\mu +\int vd\nu$ over integrable Kantorovich potentials satisfying $u(x)+v(y)\le c(x,y)$. The primal minimum is attained; a dual maximum needs additional assumptions. Compact metric spaces and a finite continuous cost suffice for attainment of both extrema.

Lebesgue measure on the real numbers extends interval length: $\lambda ([a,b])=b-a$ for $a\le b$. Its restriction to the Borel sigma-algebra is the unique translation-invariant sigma-finite measure on that sigma-algebra normalized by $\lambda ((0,1])=1$. The full Lebesgue measure is the completion of a measure of this restriction; its domain is the Lebesgue measurable sets. Translation invariance follows by comparing a translated Borel measure on half-open intervals and applying the uniqueness theorem for measures. For the construction and interval properties, see MIT lecture 8.

Every singleton has Lebesgue measure zero: $\{x\}\subseteq [x-\epsilon ,x+\epsilon ]$ gives $0\le \lambda (\{x\})\le 2\epsilon$ for every $\epsilon >0$. Enumerating the countable set $\mathbb{Q}\cap [0,1]$ and applying countable additivity yieldsThus a dense set can be a null set, and a union of null sets indexed by an uncountable set can have positive Lebesgue measure. The distinction explains both the Dirichlet function example and why almost everywhere is weaker than everywhere.

For a positive Borel measure on ${\mathbb{R}}^d$, let $E_{\alpha }=\{M_u\nu >\alpha \}$, using open balls in the uncentered maximal function of a finite measure. Every ball witnessing the inequality lies entirely in $E_{\alpha }$. Cover a compact subset by finitely many witnessing balls and select disjoint balls greedily by decreasing radius. Their threefold dilations cover the compact set. Their total measure is at most $\nu (E_{\alpha })$, giving the bound by inner regularity of Lebesgue measure. For a signed or complex measure, apply this to its total variation measure.

Let $A\subseteq {\mathbb{R}}^d$ be measurable and let $\mu$ be a locally finite positive Borel measure. At a point in the support of a measure of $\mu$, all sufficiently small balls have finite positive mass. Its relative density iswhen the limit exists. The ratio lies in $[0,1]$. The support condition prevents zero denominators; at a point outside the support, sufficiently small balls have no mass and this ratio is undefined without an extra convention.

For Lebesgue measure on the real line and $A=[0,1]$, the density is one at interior points, zero at exterior points and one-half at either endpoint. For $\mu ={\delta }_0$, the density at its support point is $1_A(0)$. This is a fraction of one measure's local mass. It differs from the spherical derivative of a measure, which compares the mass of a measure with Euclidean volume. The Lebesgue density theorem gives the almost-everywhere zero-or-one statement when the reference is Lebesgue measure.

Let $X$ be a compact metric space, let $\mu$ be a finite Borel measure whose support of a measure is all of $X$, and let $K$ be a continuous Hermitian positive-semidefinite kernel. The integral operatoris a positive compact operator on $L^2(\mu )$. Its strictly positive eigenvalues ${\lambda }_j$ have orthonormal eigenfunctions with continuous representatives ${\varphi }_j$, and Mercer's theorem givesThe series converges uniformly on $X\times X$ and absolutely at every pair of points. These measure and support assumptions specify where the eigenfunction expansion reconstructs the original kernel.^[ref]

For why full support of a measure matters, take $X=\{0,1\}$, $\mu ={\delta }_0$, and $K(1,1)=1$ with all other values zero. This continuous positive-semidefinite kernel gives the zero integral operator on $L^2(\mu )$, so its eigenfunctions cannot recover the value at the unmeasured point1. For a full-support measure, integration of the diagonal expansion instead gives ${\sum }_j{\lambda }_j={\int }_{X}K(x,x)d\mu (x)$. This connects the trace-class operator trace to the diagonal of its continuous integral kernel.

A moment problem asks whether specified numbers $m_0,m_1,\dots$ arise from a positive Borel measure $\mu$ on a prescribed subset of the real line:Existence, support restrictions and uniqueness are separate parts of the question. A probability measure must have $m_0=1$. A point mass at $a$ gives $m_n=a^n$, while the uniform probability measure on $[0,1]$ gives $m_n=1/(n+1)$.

Positivity imposes concrete constraints. For every finite coefficient vector $c$,so every associated Hankel matrix is positive semidefinite. On a fixed compact interval, the full moment sequence determines a finite measure uniquely: equality on polynomials extends to equality on continuous functions by the Weierstrass approximation theorem. A finite list of moments generally leaves many possible measures, which explains the distinction between moment reconstruction and finite-data fitting.

A non-atomic measure contains no atom of a measure. For a finite Borel measure on a Polish space, this is equivalent to giving every singleton mass zero. For a probability measure on $\mathbb{R}$, it is also equivalent to continuity of its cumulative distribution function.

Put $K=\sigma (T)$ and use the continuous functional calculus $\pi :C(K)\to \mathcal{B}(H)$. It is a unital star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^{\ast }(I,T,T^{\ast })$, together with spectral permanence for C-star subalgebras, which is among the permitted C-star results.

Take the Hilbert space inner product linear in its first variable. For a vector $\xi$, let $H_{\xi }=\overline{\{\pi (f)\xi :f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi (f)$ and its adjoint, so it is reducing. The functional $f\mapsto ⟨\pi (f)\xi ,\xi ⟩$ is positive because $f\ge 0$ implies $\pi (f)=\pi (\sqrt{f}{)}^{\ast }\pi (\sqrt{f})$. It has, by the Riesz-Markov-Kakutani representation theorem, a finite regular positive Borel measure ${\mu }_{\xi }$ on $K$. The identitymakes $f\mapsto \pi (f)\xi$ an isometry on its quotient by the functions of zero $L^2$ norm. Density of continuous functions in $L^2({\mu }_{\xi })$ extends it to a unitary $U_{\xi }:L^2({\mu }_{\xi })\to H_{\xi }$. Multiplication by a continuous function intertwines with $\pi$ on this subspace.

Choose a maximal family of mutually orthogonal nonzero cyclic reducing subspaces. Their orthogonal sum is all of $H$: otherwise its reducing orthogonal complement contains a nonzero vector generating another such subspace. This argument does not require separability. For a bounded Borel measurable function $g$ on $K$, multiplication $M_g$ on every $L^2({\mu }_{\xi })$ is bounded with norm at most $\Vert g{\Vert }_{\infty }$. Define on that orthogonal sumMultiplication operators satisfy $M_{fg}=M_f{M}_g$, $M_{\overline{g}}=M_g^{\ast }$ and $M_1=I$. The same identities hold for their direct sums, and the norm bound is uniform in $\xi$. Thus ${\beta }_T$ is a norm-decreasing unital star-homomorphism on the algebra of actual bounded Borel measurable functions, and ${\beta }_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.

Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ for $z\ne 0$, taking $\mathrm{Arg}\in [0,2\pi )$. Let $S={\beta }_T(s)$. ThenThus every bounded normal operator has a normal square root. The Borel branch is essential in general: on a circle winding about zero a continuous square-root branch need not exist.

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

For a positive Borel measure $\nu$, the centered Hardy-Littlewood maximal function and the uncentered maximal function of a finite measure areWe use open balls and allow infinite values. For signed or complex measures the definition uses their total variation measure. Since a centered ball is also an admissible uncentered ball, and every ball $B(c,r)$ containing $x$ lies in $B(x,2r)$,For functions, $m(f)$ means the centered Hardy-Littlewood maximal function of $|f|dx$.

The set $E_{\alpha }$ is the union of the open balls $B$ with $\nu (B)>\alpha |B|$. In particular, each witnessing ball lies entirely inside $E_{\alpha }$. If $K\subset E_{\alpha }$ is compact, take a finite witnessing cover of $K$. Choose its largest-radius ball, discard all balls meeting it, and repeat. The selected balls $B_j$ are disjoint. Every discarded ball has no larger radius than the selected ball that discarded it and is contained in its threefold dilation. HenceTaking the supremum over compact $K$ by inner regularity of Lebesgue measure proves the stronger, localized estimateIf the right side is infinite there is nothing to prove. Positivity of the measure is important: a signed version uses $|\nu |(E_{\alpha })$, not $\nu (E_{\alpha })$.

For the radial-kernel inequality, integrability implies $\varphi (r)\to 0$. For $0<s<\varphi (0)$, let $r(s)$ be the unique radius with $\varphi (r(s))=s$. The layer cake representation and Tonelli theorem giveThe last integral equals the total mass of $\varphi (|\cdot |)$, namely one. Taking the absolute value of the original convolution therefore proves the claim with constant exactly one. This is radial decreasing kernel domination by the maximal function.

An application is boundary convergence of the Poisson integral. Its normalized radial profile is $c_d(1+r^2{)}^{-(d+1)/2}$, so ${\sup }_{t>0}|P_t\ast f|\le m(f)$. For a continuous compactly supported $g$, the approximate identity property gives $P_t\ast g\to g$ pointwise. For arbitrary $f\in L^1$, put $h=f-g$; thenThe preceding weak type (1,1) estimate and the elementary integral bound for $|h|$ show that the set where this exceeds $2a$ has measure at most $(3^d+1)\Vert h{\Vert }_1/a$. Continuous compactly supported functions are dense in $L^1$, so this measure is zero. Thus the Poisson integrals of an integrable function converge to it almost everywhere; the same approximate identity also gives convergence in $L^1$.