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.

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.

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.

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.

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

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.

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 $A$.

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 $n$.

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:

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.

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.

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.

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 $\xi$.

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.

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 $\Vert L\Vert =L(1)=\mu (\mathbb{T})$.

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.

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.