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The complement of a comeagre set is a meagre set, a countable union of nowhere dense sets. A dense G-delta set is comeagre. The Baire category theorem ensures that two comeagre subsets of a nonempty complete metric space cannot be disjoint.

A subspace of a complete metric space admits a compatible complete metric if and only if it is a G-delta set. For the reverse direction, adjoining the coordinates $1/d(x,X\setminus U_n)$ to the ambient metric prevents Cauchy sequences from approaching any excluded closed boundary. For the forward direction, small-diameter relative open covers in a compatible complete metric produce the ambient open sets.

A subspace is topologically complete if its subspace topology is induced by some complete metric. The compatible complete metric need not be the restriction of the displayed ambient metric. For example, an open interval is topologically complete although its usual metric is incomplete.

Suppose $\rho$ is a compatible complete metric on $Y$. For each $n\ge 1$ and $y\in Y$, choose an ambient open set $V_{n,y}$ containing $y$, contained in $B_d(y,1/n)$, and with $\rho$-diameter of $V_{n,y}\cap Y$ at most $1/n$. Such sets exist because the two topologies on $Y$ agree. Put $U_n={\bigcup }_{y\in Y}{V}_{n,y}$. Clearly $Y\subseteq {\bigcap }_n{U}_n$.

If $x\in {\bigcap }_n{U}_n$, choose $y_n\in Y$ with $x\in V_{n,y_n}$. Then $d(x,y_n)<1/n$, so $y_n\to x$ in the ambient metric. For fixed $m$, the point $x$ belongs to some open $V_{m,z}$, and eventually all $y_n$ belong to this set. Their pairwise $\rho$-distances are then at most $1/m$. Hence $(y_n)$ is $\rho$-Cauchy, and completeness gives a limit $y\in Y$. Compatibility gives $d(y_n,y)\to 0$, forcing $x=y$. ThusThis proves the G-delta criterion for topological completeness in the required direction.

The converse is true in a complete metric ambient space. If $Y={\bigcap }_n{U}_n$ with each $U_n$ open in complete $X$, define $a_n(y)=1/d(y,X\setminus U_n)$, omitting any term whose complement is empty. A compatible complete metric on $Y$ isContinuity of each $a_n$ and the uniformly small series tail show that $D$ has the original topology. A $D$-Cauchy sequence is $d$-Cauchy, hence converges to some $x\in X$. For each $n$, its real coordinates $a_n$ form a Cauchy sequence and remain bounded. Since distance to a closed set is continuous, $d(x,X\setminus U_n)>0$, so $x\in U_n$ for every $n$. The same coordinate convergence and series-tail argument give convergence in $D$. Thus $D$ is complete.

For the normed-space claim, embed $E$ densely in its norm completion $\hat{E}$. Topological completeness and the preceding criterion make $E$ a dense G-delta set in $\hat{E}$, hence comeagre. Every translate $x+E$ is also comeagre. The Baire category theorem in the complete space $\hat{E}$ implies that $E\cap (x+E)$ is nonempty. If $z=x+e$ lies in that intersection, then $x=z-e\in E$. This holds for every $x\in \hat{E}$, soThis is the comeagre subgroup completeness argument.

Finally let $B=\{\ell \in E^{\ast }:\Vert \ell \Vert <1\}$ with its weak-star topology. Put $r_n=1-1/(n+1)$ and $F_n=\{\ell \in B:\Vert \ell \Vert \le r_n\}$. The dual norm is the supremum of the continuous evaluation moduli on the unit ball of $E$, so every $F_n$ is relatively closed. Also $B={\bigcup }_n{F}_n$.

Each $F_n$ has empty relative interior. Given $\ell \in F_n$ and a basic weak-star neighborhood restricting finitely many evaluations on $x_1,\dots ,x_k$, the Hahn-Banach theorem gives a nonzero functional $h$ annihilating their finite-dimensional span, because $E$ is infinite-dimensional. All $\ell +th$ retain those evaluations. Their norm varies continuously with $t$ and becomes unbounded; choose $t$ so that $r_n<\Vert \ell +th\Vert <1$. This point lies in the neighborhood inside $B$ but outside $F_n$. Thus $B$ is a nonempty countable union of relatively closed nowhere dense sets and is not a Baire space. ConsequentlyThis weak-star open dual ball category obstruction works without assuming that $E$ is separable or that the weak-star ball is metrizable.

The weak topology of probability measures is the smallest topology making every map $\mu \mapsto \int hd\mu$ continuous for bounded continuous real $h$ on $X$. Define the bounded-Lipschitz metric byUsing the maximum of the two bounds instead of their sum changes the metric by at most a factor of two and gives the same topology. Symmetry and the triangle inequality follow from the supremum formula. Bounded Lipschitz functions distinguish probability measures by the open-set approximations below, so $\beta$ is a metric. On a separable metric space, $\beta$ induces the weak topology.

Here are the essential details. If $\beta ({\mu }_n,\mu )\to 0$, integrals of every bounded Lipschitz function converge after rescaling its Lipschitz bound. For an open set $A\ne X$, the functionsincrease to $1_A$; for $A=X$ take $h_m=1$. Thus $\mu (A)={\sup }_m\int h_{m}d\mu$. It follows both that $\mu \mapsto \mu (A)$ is lower semicontinuous for the weak topology, as a supremum of continuous maps, and that beta convergence impliesThis is the open-set Portmanteau criterion. Its closed-set equivalent, by complements, isEither family of inequalities is necessary and sufficient for weak convergence of probability measures. For sufficiency of the open inequalities, apply the layer-cake formula and Fatou's lemma to the open superlevel sets of a continuous function $0\le h\le 1$, obtaining $\mathrm{liminf}\int hd{\mu }_n\ge \int hd\mu$. Applying the same argument to $1-h$ gives the reverse bound. Scaling handles every bounded continuous function.

Conversely, weak convergence implies beta convergence. Choose a compact set $K$ of arbitrarily large $\mu$-mass, using tightness of a probability measure on a Polish space. The unit bounded-Lipschitz class is uniformly bounded and equicontinuous on $K$, so finitely many of its members approximate all others uniformly there. On the open $\delta$-neighborhood $K^{\delta }$, approximation errors increase by at most $2\delta$. The open-set inequality makes ${\mu }_n(K^{\delta })$ large for all sufficiently large $n$. Weak convergence for the finitely many selected functions, combined with these approximation errors and the small mass outside $K^{\delta }$, bounds the supremum defining $\beta$ by an arbitrarily small number. The same finite-test and open-mass bounds define weak neighborhoods, so the argument gives equality of the topologies, not only their convergent sequences. It does not first assume uniform tightness of the entire sequence.

The map $j=i_{\ast }$ is injective and continuous: for $h\in C(\tilde{X})$, $h\circ i$ is bounded continuous on $X$, and $\int hdj(\mu )=\int h\circ id\mu$. To prove inverse continuity on its image, suppose $j({\mu }_n)\to j(\mu )$. For every open $A\subset X$, there is an open $U\subset \tilde{X}$ with $i(A)=U\cap i(X)$. Hence ${\mu }_n(A)=j({\mu }_n)(U)$ and $\mu (A)=j(\mu )(U)$. The open-set criterion on $\tilde{X}$ now gives the criterion on $X$, so ${\mu }_n\to \mu$. Both spaces are metrizable, and this sequential argument provesThis is the probability pushforward embedding theorem. The printed claim needs the words “onto its image”: it is generally not onto all $P(\tilde{X})$, since a Dirac mass at a point outside $i(X)$ is not in its image. By Question 1, the completely metrizable subspace $i(X)$ is a Borel G-delta set, and

Now let $S$ be weakly closed and uniformly tight. Take any sequence $({\mu }_n)$ in $S$. The space $P(\tilde{X})$ is compact for the weak topology. For completeness, this compactness follows by choosing a countable uniformly dense family in $C(\tilde{X})$, extracting a diagonal subsequence of its bounded integrals, and extending the limits to a positive normalized functional on $C(\tilde{X})$; the Riesz-Markov-Kakutani representation theorem gives the limiting probability measure. Thus, along a subsequence, $j({\mu }_n)\to \lambda$.

For every $m$, choose compact $K_m\subset X$ with $\mu (K_m)\ge 1-2^{-m}$ for every $\mu \in S$. The set $i(K_m)$ is compact and closed in $\tilde{X}$, so the closed-set criterion givesTherefore $\lambda (i(X))=1$, and $\lambda =j(\mu )$ for some $\mu \in P(X)$. Inverse continuity gives ${\mu }_n\to \mu$, and closedness puts $\mu \in S$. Hence $S$ is sequentially compact, and metrizability makes it compact. This proves the required closed uniformly tight compactness criterion, rather than assuming it from Prokhorov's theorem.

Such a compact metrizable extension is always available: for a countable dense set $(x_k)$, the coordinates $d(x,x_k)/(1+d(x,x_k))$ embed $X$ homeomorphically into $[0,1{]}^{\mathbb{N}}$. Injectivity and inverse continuity follow by choosing $x_k$ close to a specified point and using the triangle inequality. The closure of the image gives $\tilde{X}$.

To show separability of $P(X)$, use finite sums of Dirac measures on a countable dense subset of $X$ with nonnegative rational weights summing to one. This is a countable family. Given $\mu$, cover a large-mass compact set by finitely many small balls centered in that subset, move the mass in each piece to its center, and move the remaining small mass to one fixed center. The bounded-Lipschitz error is at most the ball radius plus twice the exceptional mass. Approximate the resulting weights by rational weights. Thus these atomic measures are beta-dense.

Finally use a complete compatible metric $d$ on the Polish space $X$, as required for the granted total-boundedness-to-tightness assertion. A beta-Cauchy sequence is beta-totally bounded, and so is its closure in $P(X)$. By the allowed assertion this closure is uniformly tight; it is weakly closed because the weak and beta topologies agree. The compactness result just proved gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge in beta. Consequently beta is complete for this choice of $d$, andIf the originally displayed compatible metric is incomplete, replace it by a complete compatible metric for this last argument; the weak topology itself is unchanged.

Use the standard setting of a compact metrizable convex set $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real functions on $K$. The affine upper envelope of a bounded real function isConstants make this infimum finite, and $f\le \overline{f}$. An infimum of affine continuous majorants is concave and upper semicontinuous, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric measures; it is not the pointwise maximum of $f$ and a selected affine function.

For a fixed probability $\mu$, define $p(g)={\int }_K\overline{g}d\mu$ on real $C(K)$. Affine majorants showThus $p$ is sublinear and $p(a)=\int ad\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The Hahn-Banach theorem extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.

If $g\ge 0$, then $p(-g)\le 0$, so $L(g)\ge 0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le \Vert g{\Vert }_{\infty }$, and the Riesz-Markov-Kakutani representation theorem produces a Borel probability $\nu$ with $L(g)=\int gd\nu$. ThereforeFor affine $a$, testing both $a$ and $-a$ gives $\int ad\nu =\int ad\mu$: the two measures have the same barycenter. This proves the requested supporting measure lemma for affine upper envelopes.

Choquet's theorem: every $x\in K$ has a Borel probability measure $\lambda$ concentrated on the extreme points of $K$ whose barycenter is $x$; equivalently,Concentration is a measure-one assertion, not a claim that the topological support must be a closed subset of $\mathrm{Ex}K$.

To prove it, let ${\mathcal{M}}_x$ be the measures satisfying all the displayed affine equalities. It is nonempty because it contains ${\delta }_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda \in {\mathcal{M}}_x$ maximizing $\int hd\lambda$. Apply the supporting measure lemma with $\mu =\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu \in {\mathcal{M}}_x$, andThus $\overline{h}-h\ge 0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant giveTherefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\mathrm{Ex}K$ is Borel by the allowed G-delta set assertion. Its zero integral forces $\lambda (K\setminus \mathrm{Ex}K)=0$, proving Choquet's theorem by strict convexity. No uniqueness is asserted for the representing measure.

For the real $L^{\infty }$ example, give its closed unit ball $K$ the weak-star topology $\sigma (L^{\infty },L^1)$, not the norm topology. The Banach-Alaoglu theorem makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le 1-\epsilon$ on a positive-measure set, adding and subtracting $\epsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le 1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the extreme-point criterion for the L-infinity unit ball.

For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus (A\cup B)$ and $h_{\pm }=1_A-1_B\pm 1_C$. Then $h_{\pm }$ are extreme andhas barycenter $f$. If $C$ is null, the two point masses coincide.

For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le 1$,Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the function $t\mapsto \int u(s)H_t(s)ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The measure is independent of changes to $f$ on a null set. For each $s$,The integrand paired with $u$ is dominated by $|u|$, so Fubini's theorem yieldsThese continuous linear tests define the weak-star barycenter, hence that barycenter is $f$. Together with $\nu (\mathrm{Ex}K)=1$, this is the required threshold Choquet representation in L-infinity.

If $L^{\infty }$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta (s)$, with $r=|f|$ and $|\zeta |=1$, choosing $\zeta =1$ where $f=0$. Replace the threshold family by $\zeta (s)(21_{\{t\le (1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing measure in the real locally convex interpretation of the complex ball.