Weak Convergence Tightness and Representation

(i) implies (ii) because a uniformly continuous function is continuous. Suppose (ii), and let F be nonempty and closed. By F1, $f_m(x)=\max (0,1-md(x,F))$ is bounded and uniformly continuous. Moreover $f_m\downarrow {\mathbf{1}}_F$. Thus ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(F)\le \int f_{m}d\mu$ for every m. F2 with majorant one gives (iii) as m tends to infinity. For F empty the inequality is zero<=zero.

For G open, apply (iii) to its closed complement and use ${\mu }_n(G)=1-{\mu }_n(S\setminus G)$ to obtain (iv). Conversely the same complement calculation obtains (iii) from (iv). If A is a Borel continuity set, $A^{\circ }\subseteq A\subseteq \bar{A}$ and $\mu (A^{\circ })=\mu (A)=\mu (\bar{A})$. The open lower bound and closed upper bound therefore squeeze ${\mu }_n(A)$ to $\mu (A)$, proving (v).

Assume (v), and fix a bounded continuous real f and $\eta$>0. The disjoint level sets with $\mu (f=t)\ge 1\mathrm{/}r$ number at most r for each positive integer r. Their union over r contains all positive-mass levels and is countable (each finite subset of the real line can be listed in increasing order). Choose finitely many increasing levels $t_0<\cdots <t_m$ outside this countable set, with $t_0<-\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, $t_m>\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and mesh below $\eta$. Such levels exist in every open interval, since an interval is uncountable.

For $A_j=\{t_{j-1}\le f<t_j\}$, continuity of f gives $\mathrm{\partial }{A}_j\subseteq \{f=t_{j-1}\}\cup \{f=t_j\}$, so (v) applies. The simple function $s={\sum }_j{t}_{j-1}{\mathbf{1}}_{A_j}$ satisfies $\mathrm{\mid }f-s\mathrm{\mid }\le \eta$ everywhere. Therefore $\mathrm{\mid }\int fd{\mu }_n-\int fd\mu \mathrm{\mid }\le 2\eta +\mathrm{\mid }{\sum }_j{t}_{j-1}({\mu }_n(A_j)-\mu (A_j))\mathrm{\mid }$. The finite sum tends to zero, by step 1.3 and (v). Letting $\eta$ tend to zero proves (i), closing all equivalences.

Use F4 for increasing rays. Use F3 for decreasing rays and intervals of finite probability. For a probability law $\mu$, write $F(t)=\mu ((-\mathrm{\infty },t])$. Continuity from above and below of finite measures show F is right-continuous, has limits zero and one at the two infinities, and has jump $\mu (\{t\})$ at t. Thus a continuity point has $\mu (\{t\})=0$. If ${\mu }_n\Rightarrow \mu$, F1 on $(-\mathrm{\infty },t]$ gives $F_n(t)\to F(t)$ at every such point, exactly F2.

Conversely assume convergence of CDFs at continuity points of F. Fix bounded continuous f, $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\eta$>0. Choose continuity points a<b with $\mu ((-\mathrm{\infty },a])+\mu ((b,\mathrm{\infty }))<\eta$. They exist because tails tend to zero and the positive jumps form a countable set: at most r atoms have mass at least 1/r. CDF convergence makes the same sum of two tails less than 2eta for all large n.

The closed bounded interval $[a,b]$ is compact by F5. On [a,b], continuity is uniform: for each point choose a neighborhood on which oscillation is small, extract a finite subcover by compactness, and use the minimum of the finitely many smaller radii. Choose a finite partition $a=t_0<\cdots <t_m=b$ by continuity points with oscillation of f on each interval below $\eta$. Then ${\mu }_n((t_{j-1},t_j])=F_n(t_j)-F_n(t_{j-1})$ converges to the corresponding $\mu$ mass. Integrals of the finite step approximation therefore converge. Its error inside (a,b] is at most $\eta$ for each law; the outside error is at most $3M\eta$ in the comparison of the two integrals, by step 1.2. Let $\eta$ decrease to zero. This proves weak convergence.

If $\mu$ and $\nu$ have equal integrals of every bounded continuous test, the constant sequence $\mu$ converges weakly to $\nu$. F1 gives $\mu (F)\le \nu (F)$ for every closed F. Reverse the roles to get equality.

The class of Borel sets on which the two probabilities agree contains S, is closed under complements and disjoint countable unions, and contains the closed sets by step 1.1. Closed sets form a $\pi$-system generating the Borel $\sigma$-algebra; F2 therefore gives equality on all Borel sets. If a sequence has two weak limits, uniqueness of each numerical integral limit gives the hypothesis of step 1.1, so those limits agree.

For $r\ge 1$ let $U_r$ be the union of all open subsets V of S such that the diameter of g(V) is less than 1/r. The continuity set is ${\bigcap }_r{U}_r$: continuity at x supplies such a neighborhood by making all image points within 1/(3r) of g(x); conversely a neighborhood with image diameter below $\epsilon$ forces $d(g(y),g(x))<\epsilon$ there. Thus $D_g$ is Borel.

If F is closed in T and x lies outside $g^{-1}(F)\cup D_g$, continuity at x and the open complement of F give a neighborhood disjoint from $g^{-1}(F)$. Hence $\overline{g^{-1}(F)}\subseteq g^{-1}(F)\cup D_g$. Applying F1 to this closed preimage closure gives ${\mathrm{lim\hspace{0.17em}sup}}_n{\mu }_n(g^{-1}(F))\le \mu (\overline{g^{-1}(F)})\le \mu (g^{-1}(F))$.

The inequality in step 1.2 is the closed-set bound for the pushforward probabilities, so F1 gives their weak convergence. F2 identifies these pushforwards with the laws of g($X_n$) and g(X), proving the random-element formulation.

By F3, distance to nonempty $F$ is continuous; its sublevel sets are closed. For nonempty closed F put $F^{[\epsilon ]}=\{x:d(x,F)\le \epsilon \}$, which is closed. If $Y_n$ lies in F and $d(X_n,Y_n)\le \epsilon$, then $X_n$ lies in this enlargement. Thus $P(Y_n\in F)\le P(X_n\in F^{[\epsilon ]})+P(d(X_n,Y_n)>\epsilon )$.

F1 and the probability hypothesis give ${\mathrm{lim\hspace{0.17em}sup}}_{n}P(Y_n\in F)\le P_X(F^{[\epsilon ]})$. For $m\in \mathbb{N}$, the sets $E_m:=F^{[1\mathrm{/}(m+1)]}$ decrease to F, so F2 makes their probabilities decrease to P_X(F). Empty F has probability zero without an enlargement. The resulting closed-set bound is again F1, now proving $Y_n\Rightarrow X$.

AC restricted to any countable nonempty family gives countable choice. Thus F1 applies to the given Polish S and its probability $\mu$. Take the Borel set A=S; for each $\epsilon$>0 it supplies compact K with $\mu (S\setminus K)<\epsilon$. This is F2 for the one-law family.

The complete totally bounded criterion F5 applies under countable choice, already supplied by AC in step 1.1. Use F4 on the countably many omitted sets. Use F3 on the increasing finite unions below. The compact-set construction behind this application can be made explicit. Fix a compatible complete metric and a countable dense sequence $(a_i)$. For each $m\ge 1$, finite initial unions of open balls $B(a_i,2^{-m})$ increase to S; choose their least length with loss below $\epsilon 2^{-m-1}$. Let $C_m$ be the corresponding finite union of closed balls, and $K={\bigcap }_m{C}_m$. Subadditivity gives $\mu (S\setminus K)\le {\sum }_m\mu (S\setminus C_m)<\epsilon$. K is closed and hence complete. For any $\eta$>0 choose m with $2^{1-m}<\eta$; each selected ball meeting K contributes one point of K, and those finitely many points form an $\eta$-net in K. Thus K is totally bounded and complete, hence compact, which realizes the bound in step 1.1.

If K is empty there is one function and the assertion holds. Otherwise F1 supplies finite 1/m-nets. AC chooses these nets with finite listings; their countable union D is dense. Consider all functions $x\mapsto {\min }_{1\le j\le l}(q_j+Ld(x,a_j))$, with finite lists $a_j\in D$, rational $q_j$ and positive integer L, optionally clipped between rational constants. These form a countable family of continuous functions; F2 with singleton sets gives the needed continuity.

Fix continuous f, $\eta$>0 and $M\ge \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. By F3 choose $\delta$>0 so $d(x,y)<\delta$ implies $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\eta$. Choose integer L with $L\delta >2M$. Then $g(x)={\inf }_{y\in K}(f(y)+Ld(x,y))$ obeys $g(x)\le f(x)$ by y=x. For d(x,y)<$\delta$ the expression is at least f(x)-$\eta$; for d(x,y)>=$\delta$ it is greater than -M+2M>=f(x). Thus $f(x)-\eta \le g(x)\le f(x)$.

Choose a finite net $a_1,\dots ,a_l$ from D with mesh h<$\delta$ and Lh<$\eta$, and rational $q_j$ with $\mathrm{\mid }{q}_j-f(a_j)\mathrm{\mid }<\eta$. For any y choose $a_j$ within h; uniform continuity gives $f(a_j)+Ld(x,a_j)\le f(y)+Ld(x,y)+2\eta$. Taking the infimum over y and allowing the rational error proves $g(x)-\eta \le {\min }_j(q_j+Ld(x,a_j))\le g(x)+3\eta$. Together with step 1.2 the error from f is at most 3eta. Rational clipping bounds containing f(K) cannot increase it. Letting $\eta$ decrease proves density.

K cannot be empty because the given laws have mass one. List a countable dense test family $f_1,f_2,\dots$ by F1. Each numerical sequence $\int f_{j}d{\mu }_n$ is bounded by $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}$. F2 supplies nested infinite subsequences along which the first j integrals converge. AC supplies these successive selections; taking the jth index of the jth subsequence gives one increasing diagonal subsequence $n_j$ with convergence for every listed test.

For any continuous f and $\eta$>0 choose a listed test h with $\mathrm{\parallel }f-h{\mathrm{\parallel }}_{\mathrm{\infty }}<\eta$. The inequality $\mathrm{\mid }\int fd{\mu }_{n_j}-\int fd{\mu }_{n_k}\mathrm{\mid }\le 2\eta +\mathrm{\mid }\int hd{\mu }_{n_j}-\int hd{\mu }_{n_k}\mathrm{\mid }$ shows the f integrals are Cauchy. Define L(f) as their finite limit. Taking limits in finite linear combinations gives linearity; nonnegative f has nonnegative integrals and hence L(f)>=0; also L(1)=1.

The compact metric space K is Hausdorff and locally compact (K itself is a compact neighborhood of each point), and $C_c(K)=C(K)$. The positive functional in step 1.2 therefore satisfies F3. Its representing Borel measure has total mass L(1)=1. The defining identity L(f)=integral f against that measure, combined with step 1.2, is weak convergence of the extracted subsequence.

The empty family satisfies both definitions vacuously. Otherwise fix a compatible complete metric on S. By F1 there is a homeomorphic embedding e into $H=[0,1{]}^{\mathbb{N}}$. H has complete metric $\rho (x,y)={\sum }_{j=0}^{\mathrm{\infty }}{2}^{-(j+1)}\mathrm{\mid }{x}_j-y_j\mathrm{\mid }$ by F2. It is totally bounded: choose an integer $N\ge 1$ with tail ${\sum }_{j\ge N}{2}^{-(j+1)}<\eta \mathrm{/}2$ and a finite mesh in coordinates $0,\dots ,N-1$ with weighted error below $\eta$/2, putting zero in later coordinates. AC restricted to countable nonempty families gives the countable choice required by F3, which makes H compact.

Assume tightness and take any sequence ${\mu }_n$ in the family. Push it forward by e. F4 gives a subsequence ${\nu }_{n_j}=e_{\ast }{\mu }_{n_j}\Rightarrow \nu$ on H. For every integer $m\ge 1$, choose compact $K_m$ in S with all ${\mu }_n$($K_m$)>1-1/m. Their images are compact and closed in H, so F5 gives $\nu (e(K_m))\ge {\mathrm{lim\hspace{0.17em}sup}}_j{\nu }_{n_j}(e(K_m))\ge 1-1\mathrm{/}m$. Hence the Borel set $E={\bigcup }_{m\ge 1}e(K_m)\subseteq e(S)$ has $\nu$ mass one.

For Borel B in S, e(B) is Borel relative to e(S). Since E is ambient Borel and contained in e(S), $E\cap e(B)$ is Borel in H. Define $\mu (B)=\nu (E\cap e(B))$; disjoint unions are preserved and $\mu$(S)=$\nu$(E)=1. For closed F in S there is a closed Z in H with $Z\cap e(S)=e(F)$, by the relative topology. Then ${\mu }_{n_j}(F)={\nu }_{n_j}(Z)$ and $\nu (Z)=\nu (Z\cap E)=\mu (F)$. The closed bound from F5 gives ${\mathrm{lim\hspace{0.17em}sup}}_j{\mu }_{n_j}(F)\le \mu (F)$, hence weak convergence on S. This proves tightness implies relative sequential compactness without assuming e(S) is Borel.

For the reverse, let $U_i$ be any countable open cover of S. Fix $\epsilon$>0. If no finite initial union works uniformly, AC selects ${\mu }_n$ in the family with ${\mu }_n({\bigcup }_{i\le n}{U}_i)\le 1-\epsilon$. Relative sequential compactness gives a weakly convergent subsequence with probability limit $\mu$. For any fixed r, eventually $n_j$>=r, so for the fixed open set $G_r={\bigcup }_{i\le r}{U}_i$ one has ${\mu }_{n_j}(G_r)\le 1-\epsilon$ eventually. The open bound in F5 yields $\mu (G_r)\le {\mathrm{lim\hspace{0.17em}inf}}_j{\mu }_{n_j}(G_r)\le 1-\epsilon$. F6 as r tends to infinity would give $\mu$(S)<=1-$\epsilon$, a contradiction. Thus a uniform finite initial union exists.

Apply step 1.4 to the dense-center ball cover of radius $2^{-m}$ and loss $\epsilon 2^{-m-1}$ at each $m\ge 1$. Let $C_m$ be the corresponding finite union of closed balls and $K={\bigcap }_m{C}_m$. F7 bounds every $\mu$(S\K) by ${\sum }_m\epsilon 2^{-m-1}<\epsilon$. K is closed in the complete S and is totally bounded: for any $\eta$ choose m with $2^{1-m}<\eta$ and select one point of K from each of the finitely many balls meeting it. These form an $\eta$-net. F3 makes K compact, proving tightness.

Consider any sequence of laws from the displayed family. If some law occurs infinitely often, it has a constant subsequence. Otherwise each law occurs only finitely often. Assign every law unequal to $\mu$ its least index in the original sequence. Removing finitely many selected terms for each bounded set of such indices leaves a subsequence whose assigned indices increase to infinity; its weak limit is $\mu$ by the given convergence.

Thus every sequence in the family has a weakly convergent subsequence with a probability limit on S. The reverse implication of F1, with the stated AC and Polish hypotheses, yields tightness.

Let $\mathcal{A}=\{{\mu }_n:n\ge 1\}$. The given uniform compact bounds are exactly tightness of this family. The forward implication of F1 makes it relatively sequentially compact.

Apply that property to the original sequence itself. It supplies increasing indices $n_j$ and a Borel probability $\mu$ on S with ${\mu }_{n_j}\Rightarrow \mu$. In particular the limit has mass one and lies on S, as asserted.

By F1, $0\le \pi \le 1$ and symmetry holds. Since $F\subseteq F^{[\epsilon ]}$, every positive $\epsilon$ is admissible for equal measures, giving self-distance zero. Admissibility is upward closed because larger radii enlarge sets and increase the error.

If $\pi$($\mu$,$\nu$)=0, for each positive integer m the upward-closure observation makes 1/m admissible. For nonempty closed F the sets $F^{[1\mathrm{/}m]}$ are closed by F2 and decrease to F. F3 gives $\mu$(F)<=$\nu$(F) and, symmetrically, the reverse. Equality also holds on the empty set. The sets on which the two probabilities agree form a lambda-system; closed sets form a generating $\pi$-system, so F4 yields $\mu$=$\nu$.

If a is admissible between $\mu$ and $\nu$, and b between $\nu$ and $\sigma$, then for every closed F, $\mu (F)\le \nu (F^{[a]})+a\le \sigma ((F^{[a]}{)}^{[b]})+a+b\le \sigma (F^{[a+b]})+a+b$. The last inclusion follows from the metric triangle inequality by approximating each infimum within any positive slack and then taking its infimum; empty F is separate. The reverse inequality interchanges $\mu$ and $\sigma$. Thus a+b is admissible, and taking a and b arbitrarily close above their infima proves the triangle inequality.

If $\epsilon$ is admissible in the all-Borel open convention, it is admissible for closed sets and closed enlargements, since $F^{\epsilon }\subseteq F^{[\epsilon ]}$. Conversely, if a is admissible in the closed convention, any Borel B satisfies $\mu (B)\le \mu (\bar{B})\le \nu ((\bar{B}{)}^{[a]})+a\le \nu (B^{a+\delta })+a+\delta$ for every $\delta$>0, because distance to B and its closure agree. Interchanging the measures gives the other inequality. Infima and arbitrary positive slack prove equality of the two conventions.

S is nonempty since $\mu$(S)=1. Fix a countable dense list $a_i$. For a fixed center, spheres at distinct radii are disjoint; at most r spheres have mass at least 1/r. Thus the radii with positive sphere mass form a countable union of finite, increasing-order lists. For each k,i, AC chooses $r_{k,i}\in (2^{-k-2},2^{-k-1})$ outside this countable exceptional set. The balls $B(a_i,r_{k,i})$ cover S by density, and each has diameter at most 2^{-k} and boundary contained in its null sphere.

At level k disjointize this ordered cover: $D_{k,i}=B(a_i,r_{k,i})\setminus {\bigcup }_{j<i}B(a_j,r_{k,j})$. These sets partition S; discard empty members. Their boundaries lie in the finite union of the first i sphere boundaries, hence are null by F1. Let $P_k$ consist of all nonempty intersections $D_{1,i_1}\cap \cdots \cap D_{k,i_k}$. These form a countable Borel partition, refine the preceding one, and have diameter at most 2^{-k}; their boundaries are again contained in finitely many null boundaries.

Every partition atom is Borel, so the $\sigma$-algebra they generate is contained in Borel(S). Conversely if U is open and x belongs to U, choose a ball about x contained in U and then k with 2^{-k} below its radius. The $P_k$ atom containing x lies in that ball, hence in U. Thus U is the union of the atoms, over countably many levels and members, that are contained in U. It lies in the generated $\sigma$-algebra, proving equality.