Proof of Portmanteau Theorem on a Metric Space

Throughout, every $f\in C_{b}(S)$ with $|f|\le M$ is integrable against every Borel probability measure $\nu$ on $S$, with $|\int_{S}f\,d\nu|\le M$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (ii)), and it is Borel, so that its superlevel sets $\{f>t\}$ and $\{f\ge t\}$ are Borel (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (i), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v)); $\nu(B)=\int_{S}\mathbf{1}_{B}\,d\nu$ for $B\in\mathcal{B}(S)$, and the integral is linear, and monotone between nonnegative functions (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), Basic Properties of the Integral of Nonnegative Measurable Maps (iii)); for $f\in C_{b}(S)$, $n\mapsto\int_{S}f\,d\mu_{n}$ is a real sequence (The Simple, Nonnegative, Integrable and Signed Integrals Are Predicative in the Measure and the Function, at Any Variable Positions (iii)); elements of $\mathbb{N}$ read in $\mathbb{R}$ are at least $1$ and keep their order (Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (ii), Natural Numbers and Factorials Read in the Reals: the Order Carries Over, One Is Least, and Factorials Are Positive, Recursive and Nondecreasing (i)), so that for $k\in\mathbb{N}$, $1/k$ is a positive real and $1/(k+1)\le1/k$; $\nu$ is monotone, subtractive, complemented and continuous from above (Basic Properties of a Measure (iii), Basic Properties of a Measure (iv), Basic Properties of a Measure (v), Basic Properties of a Measure (ix)); open and closed sets are Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (i), Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii)); for $A\subseteq S$, $\operatorname{int}A\subseteq A\subseteq\operatorname{cl}A$ with $\operatorname{cl}A$ closed and $\operatorname{int}A$ open, a set is open exactly when it equals its interior and closed exactly when it equals its closure (Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (i), Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (ii), Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iii), Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (v), Closure and Interior in a Metric Space: Containments, the Closure is Closed, the Interior is Open (iv)); and for nonempty $F\subseteq S$ and $x\in S$, $0\le\operatorname{dist}(x,F)\le d(x,a)$ for every $a\in F$, every real $r>\operatorname{dist}(x,F)$ exceeds $d(x,a)$ for some $a\in F$, and $\operatorname{dist}(\cdot,F)$ vanishes exactly on $\operatorname{cl}F$ (The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (i), The Distance to a Subset of a Metric Space: Bounds, the Lipschitz Estimate, Continuity, and Vanishing Exactly on the Closure (iv)). The proof shows (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Leftrightarrow$ (iv), then (iii) and (iv) together $\Rightarrow$ (v) $\Rightarrow$ (iii) $\Rightarrow$ (i).

Neighbourhoods of a closed set. Let $F\subseteq S$ be nonempty and closed. For real $\delta>0$ let

$$N_{\delta}(F)=\{x\in S:d(x,a)<\delta\text{ for some }a\in F\}.$$

This set is open: if $d(x,a)<\delta$ with $a\in F$, then every $y$ with $d(y,x)<\delta-d(x,a)$ has $d(y,a)\le d(y,x)+d(x,a)<\delta$. Choose reals $s_{j}$ with $0<s_{j}<1/j$ for $j\in\mathbb{N}$; for instance, the levels that A Measurable Real Function Has Null Level Sets at Levels Arbitrarily Close to Any Point, for a Finite Measure (ii) gives for $\mu$, the constant map $0$ (which lies in $C_{b}(S)$ by Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii), hence is Borel) and $a=0$, of which only the bounds are used. The sets

$$U_{k}=\bigcup_{j\ge k}N_{s_{j}}(F)\qquad(k\in\mathbb{N})$$

are open, contain $F$ (each $a\in F$ has $d(a,a)=0<s_{j}$) and decrease. Their intersection is $F$: if $x\in U_{k}$ for every $k$, then for every $n\in\mathbb{N}$ there are $j\ge n$ and $a\in F$ with $d(x,a)<s_{j}<1/j\le1/n$, so $\operatorname{dist}(x,F)<1/n$; as some $n$ has $1/n<\operatorname{dist}(x,F)$ whenever $\operatorname{dist}(x,F)>0$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)), $\operatorname{dist}(x,F)=0$ and $x\in\operatorname{cl}F=F$. Since $\mu(U_{1})\le1$, continuity from above gives $\mu(F)=\inf_{k}\mu(U_{k})$, the greatest lower bound of the numbers $\mu(U_{k})$ in $[-\infty,\infty]$ (Order and Suprema in the Extended Real Line (iii)). Given a real $\varepsilon>0$, $\mu(F)+\varepsilon$ is therefore not a lower bound: some $k$ has $\mu(U_{k})<\mu(F)+\varepsilon$, and $\delta=s_{k}$ satisfies $N_{\delta}(F)\subseteq U_{k}$. Thus

$$\text{for every real }\varepsilon>0\text{ there is a real }\delta>0\text{ with }\mu(N_{\delta}(F))<\mu(F)+\varepsilon.$$

Step 1 ((i) $\Rightarrow$ (ii)). A bounded Lipschitz function lies in $C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi), Bounded Continuous Functions on a Metric Space (i)); apply Weak Convergence of Finite Borel Measures on a Metric Space (i).

Step 2 ((ii) $\Rightarrow$ (iii)). The case $F=\emptyset$ is trivial. Let $F$ be nonempty and closed, and let $\varepsilon>0$ be real. Take $\delta$ as above, and $k\in\mathbb{N}$ with $1/k<\delta$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)). Let $f_{k}$ be the function of Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vii), which is bounded Lipschitz and so lies in $C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi)). We have $0\le f_{k}\le1$ and $f_{k}=1$ on $F$; and if $f_{k}(x)>0$, then $k\operatorname{dist}(x,F)<1$, so some $a\in F$ has $k\,d(x,a)<1$, whence $d(x,a)<1/k<\delta$ and $x\in N_{\delta}(F)$. Hence

$$\mathbf{1}_{F}\le f_{k}\le\mathbf{1}_{N_{\delta}(F)}.$$

Then $\mu_{n}(F)\le\int_{S}f_{k}\,d\mu_{n}$, and by (ii) the sequence $\bigl(\int_{S}f_{k}\,d\mu_{n}\bigr)_{n}$ converges to $\int_{S}f_{k}\,d\mu$, so it is bounded (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (iii)); Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (iii) and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (xi) give

$$\limsup_{n\to\infty}\mu_{n}(F)\le\int_{S}f_{k}\,d\mu\le\mu(N_{\delta}(F))<\mu(F)+\varepsilon.$$

As $\varepsilon$ was arbitrary, $\limsup_{n}\mu_{n}(F)\le\mu(F)$.

Step 3 ((iii) $\Leftrightarrow$ (iv)). First, if $(a_{n})$ and $(b_{n})$ are bounded real sequences with $a_{n}+b_{n}=1$ for every $n$, then

$$\liminf_{n\to\infty}a_{n}+\limsup_{n\to\infty}b_{n}=1.$$

Indeed, let $\varepsilon>0$ be real. By Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (v) there is $N$ with $b_{m}<\limsup_{n}b_{n}+\varepsilon/2$ for $m\ge N$, and by Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (vi) some $m\ge N$ has $a_{m}<\liminf_{n}a_{n}+\varepsilon/2$; so $1=a_{m}+b_{m}<\liminf_{n}a_{n}+\limsup_{n}b_{n}+\varepsilon$. Symmetrically, Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (v) gives $N'$ with $a_{m}>\liminf_{n}a_{n}-\varepsilon/2$ for $m\ge N'$, and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (vi) some $m\ge N'$ with $b_{m}>\limsup_{n}b_{n}-\varepsilon/2$, so $1>\liminf_{n}a_{n}+\limsup_{n}b_{n}-\varepsilon$. As $\varepsilon$ was arbitrary, the identity follows. Now $G$ is open exactly when $F=S\setminus G$ is closed (Closed Subset of a Topological Space (i)), and then $\mu_{n}(G)+\mu_{n}(F)=1$ and $\mu(G)=1-\mu(F)$; so $\liminf_{n}\mu_{n}(G)=1-\limsup_{n}\mu_{n}(F)$, and $\liminf_{n}\mu_{n}(G)\ge\mu(G)$ holds exactly when $\limsup_{n}\mu_{n}(F)\le\mu(F)$.

Step 4 ((iii) and (iv) $\Rightarrow$ (v)). Let $A\in\mathcal{B}(S)$ with $\mu(\partial A)=0$. Since $\partial A=\operatorname{cl}A\setminus\operatorname{int}A$ (Boundary of a Subset of a Metric Space (i)) with $\operatorname{int}A\subseteq\operatorname{cl}A$ both Borel, $\mu(\operatorname{cl}A)=\mu(\operatorname{int}A)+\mu(\partial A)=\mu(\operatorname{int}A)$, and so $\mu(\operatorname{int}A)=\mu(A)=\mu(\operatorname{cl}A)$. Then Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (iii), (iii) and (iv) give

$$\limsup_{n\to\infty}\mu_{n}(A)\le\limsup_{n\to\infty}\mu_{n}(\operatorname{cl}A)\le\mu(A)\le\liminf_{n\to\infty}\mu_{n}(\operatorname{int}A)\le\liminf_{n\to\infty}\mu_{n}(A),$$

and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (ii) and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (xi) give $\mu_{n}(A)\to\mu(A)$.

Step 5 ((v) $\Rightarrow$ (iii)). If $F=\emptyset$, then $\mu_{n}(F)=0=\mu(F)$ for every $n$. Let $F$ be nonempty and closed, let $\varepsilon>0$ be real, and take $\delta$, $k$ and $f_{k}$ as in Step 2. Since $f_{k}$ is Borel and $\mu$ is finite, A Measurable Real Function Has Null Level Sets at Levels Arbitrarily Close to Any Point, for a Finite Measure (i) gives a level $\tau$ with $0<\tau<1$ and $\mu(\{f_{k}=\tau\})=0$. Let

$$H=\{x\in S:f_{k}(x)>\tau\}.$$

As $f_{k}$ is Lipschitz with constant $k$, $|f_{k}(x)-f_{k}(y)|\le k\,d(x,y)$; so $H$ contains the ball of radius $(f_{k}(x)-\tau)/k$ about each of its points $x$, and is open. If $x\in\operatorname{cl}H$, then for every real $r>0$ some $y\in H$ has $d(x,y)<r$ (Closure of a Subset of a Metric Space (ii)), so $f_{k}(x)\ge f_{k}(y)-kr>\tau-kr$; hence $\operatorname{cl}H\subseteq\{f_{k}\ge\tau\}$, a Borel set. Thus the boundary $\partial H=\operatorname{cl}H\setminus H$ (Boundary of a Subset of a Metric Space (i)), a Borel set by The Boundary of a Subset of a Metric Space Is Closed and Borel (i), lies in $\{f_{k}=\tau\}$ and is $\mu$-null, and $\mu_{n}(H)\to\mu(H)$ by (v). Moreover $F\subseteq H$, as $f_{k}=1>\tau$ on $F$, and $H\subseteq N_{\delta}(F)$, as $f_{k}>0$ on $H$ (Step 2). With Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (iii) and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (xi),

$$\limsup_{n\to\infty}\mu_{n}(F)\le\mu(H)\le\mu(N_{\delta}(F))<\mu(F)+\varepsilon.$$

As $\varepsilon$ was arbitrary, $\limsup_{n}\mu_{n}(F)\le\mu(F)$.

Step 6 ((iii) $\Rightarrow$ (i)). Assume (iii). We first show, for every $f\in C_{b}(S)$, that for every real $\eta>0$ there is $N$ with

$$\int_{S}f\,d\mu_{n}\le\int_{S}f\,d\mu+\eta\qquad(n\ge N).$$

If $|f|\le M$, the function $g=(f+M+1)/(2M+2)$ lies in $C_{b}(S)$ by Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii), takes values in $[0,1]$, and $\int_{S}g\,d\nu=\bigl(\int_{S}f\,d\nu+M+1\bigr)/(2M+2)$ for every Borel probability measure $\nu$, since $\int_{S}1\,d\nu=\nu(S)=1$; so the bound for $g$ with $\eta/(2M+2)$ in place of $\eta$ gives the bound for $f$. We may therefore assume $0\le f\le1$.

For real $e>0$, let $P(e)$ be the assertion: for every $f\in C_{b}(S)$ with $0\le f\le1$, every open $G\subseteq S$ with $f=0$ on $G$, every $U\in\mathcal{B}(S)$ with $f=1$ on $U$, and every real $\eta>0$, there is $N$ with

$$\int_{S}f\,d\mu_{n}+e\,\mu(G)+e\,\mu(U)\le\int_{S}f\,d\mu+e+\eta\qquad(n\ge N).$$

$P(e)$ holds for $e\ge1$. Let $f$, $G$, $U$ and $\eta$ be as in $P(e)$. The set $S\setminus G$ is closed, and $f\le\mathbf{1}_{S\setminus G}$; by (iii) and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (v) there is $N$ with $\mu_{n}(S\setminus G)<\mu(S\setminus G)+\eta$ for $n\ge N$, so $\int_{S}f\,d\mu_{n}<1-\mu(G)+\eta$. Since $U\subseteq S\setminus G$, $\mu(G)+\mu(U)\le1$, and since $\mathbf{1}_{U}\le f$, $\mu(U)\le\int_{S}f\,d\mu$. Hence, for $n\ge N$,

$$\int_{S}f\,d\mu_{n}+e\,\mu(G)+e\,\mu(U)<1+(e-1)\bigl(\mu(G)+\mu(U)\bigr)+\mu(U)+\eta\le e+\int_{S}f\,d\mu+\eta.$$

$P(2h)$ implies $P(h)$. Let $h>0$, and let $f$, $G$, $U$ and $\eta$ be as in $P(h)$. Let $\varphi=\min(2f,1)$, which lies in $C_{b}(S)$ as the composite of $f$ with the bounded Lipschitz map $t\mapsto\max(0,\min(2t,1))$ on $\mathbb{R}$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (vi), Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iv)), let $\psi=2f-\varphi\in C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii)), and let $H=\{\varphi\ge1\}$. Then $0\le\varphi,\psi\le1$ and $\varphi+\psi=2f$; $H$ is closed, as $\varphi$ is continuous, so $S\setminus H$ is open. The triple $(\varphi,G,H)$ satisfies the hypotheses of $P(2h)$, as $\varphi=0$ on $G$ and $\varphi=1$ on $H$; so does $(\psi,S\setminus H,U)$, as $\varphi=2f$ and so $\psi=0$ on $S\setminus H$, and $\varphi=\psi=1$ on $U$. Applying $P(2h)$ to both with the same $\eta$, and taking the larger of the two indices, for all large $n$

$$\int_{S}\varphi\,d\mu_{n}+2h\,\mu(G)+2h\,\mu(H)\le\int_{S}\varphi\,d\mu+2h+\eta,\qquad\int_{S}\psi\,d\mu_{n}+2h\,\mu(S\setminus H)+2h\,\mu(U)\le\int_{S}\psi\,d\mu+2h+\eta.$$

Adding, with $\mu(H)+\mu(S\setminus H)=1$, gives $2\int_{S}f\,d\mu_{n}+2h\,\mu(G)+2h\,\mu(U)+2h\le2\int_{S}f\,d\mu+4h+2\eta$, which is the inequality of $P(h)$ multiplied by $2$.

$P(e)$ holds for every $e>0$. Suppose the set $B$ of reals $e>0$ for which $P(e)$ fails is nonempty. It is bounded above by $1$, so it has a supremum $s$ (The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (ii)), and $s>0$. As $s/2<s$, some $t\in B$ has $t>s/2$; then $2t>s$, so $2t\notin B$ and $P(2t)$ holds, whence $P(t)$ holds, contradicting $t\in B$.

Conclusion. Let $f\in C_{b}(S)$ with $0\le f\le1$ and let $\eta>0$ be real. Then $P(e)$ with $e=\eta/2$, $G=U=\emptyset$ (open and Borel, with $\mu(\emptyset)=0$) and tolerance $\eta/2$ gives $N$ with $\int_{S}f\,d\mu_{n}\le\int_{S}f\,d\mu+\eta$ for $n\ge N$, which is the bound claimed above. For any $f\in C_{b}(S)$, Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (iv) now gives $\limsup_{n}\int_{S}f\,d\mu_{n}\le\int_{S}f\,d\mu+\eta$ for every real $\eta>0$, so $\limsup_{n}\int_{S}f\,d\mu_{n}\le\int_{S}f\,d\mu$. Applied to $-f\in C_{b}(S)$ (Bounded Continuous Functions on a Metric Space: Measurability, Integrals, Operations, Lipschitz Functions and Approximation of Closed Sets (iii)), with $\int_{S}(-f)\,d\nu=-\int_{S}f\,d\nu$ and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (x), this gives $\liminf_{n}\int_{S}f\,d\mu_{n}\ge\int_{S}f\,d\mu$, and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (ii) and Properties of the Limit Superior and the Limit Inferior: Tail Formula, Order, Comparison, Bounds, Eventual and Frequent Inequalities, Subsequences, Negation and Convergence (xi) give $\int_{S}f\,d\mu_{n}\to\int_{S}f\,d\mu$. Hence $\mu_{n}\Rightarrow\mu$ by Weak Convergence of Finite Borel Measures on a Metric Space (i) (a Borel probability measure $\nu$ has $\nu(S)=1<\infty$, so it is finite by Measure, Measure Space, and Probability Measure (iii), and $(\mu_{n})$ is a sequence in the finite Borel measures on $S$, which form a set by The Measures, the Finite Measures and the Probability Measures on a Measurable Space Form Sets (ii)).