Proof of Portmanteau Theorem on a Metric Space

Throughout, $d_{\mathbb{R}}$ is the absolute-value metric on $\mathbb{R}$, and measurable means measurable with respect to $\mathcal{B}(X)$ and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line. Every bounded continuous and every bounded lower semicontinuous $f:X\to\mathbb{R}$ is measurable, by claims 3 and 5 of Borel Measurability and Bounded Integration on a Metric Space, and if $|f(x)|\le M$ for every $x$ then $f$ is integrable with respect to every probability measure $\nu$ on $(X,\mathcal{B}(X))$ with $\bigl|\int_X f\,d\nu\bigr|\le M$, by claim 6 of that lemma. In particular every sequence of integrals occurring below is bounded, so its limit inferior and limit superior are defined.

Step 0 (limits inferior and superior from eventual bounds). Let $(a_n)_{n\in\mathbb{N}}$ be a bounded real sequence, let $c$ be real, let $N\in\mathbb{N}$, and put $A_k=\{a_m:m\in\mathbb{N},\ m\ge k\}$ as in Limit Inferior of a Bounded Sequence of Real Numbers and Limit Superior of a Bounded Sequence of Real Numbers.

(i) If $c\le a_m$ for every $m\ge N$, then $c$ is a lower bound of $A_N$, so $c\le\inf A_N$; and $\inf A_N$ belongs to $\{\inf A_k:k\in\mathbb{N}\}$, whose least upper bound is $\liminf_n a_n$. Hence $c\le\liminf_n a_n$.

(ii) If $a_m\le c$ for every $m\ge N$, then $c$ is an upper bound of $A_N$, so $\sup A_N\le c$; and $\limsup_n a_n$, being the greatest lower bound of $\{\sup A_k:k\in\mathbb{N}\}$, satisfies $\limsup_n a_n\le\sup A_N\le c$.

(iii) For every real $c$, $\liminf_n(a_n+c)=\bigl(\liminf_n a_n\bigr)+c$. Indeed, if $S\subseteq\mathbb{R}$ is nonempty and bounded below and $c$ is real, write $S+c=\{s+c:s\in S\}$; then $\inf S+c$ is a lower bound of $S+c$, and every lower bound $t$ of $S+c$ makes $t-c$ a lower bound of $S$, so $t-c\le\inf S$, that is $t\le\inf S+c$; hence $\inf(S+c)=\inf S+c$. The same argument with upper bounds gives $\sup(S+c)=\sup S+c$ for nonempty $S$ bounded above. Applying the first identity to each $A_k$ gives $\inf(A_k+c)=\inf A_k+c$, and applying the second to $\{\inf A_k:k\in\mathbb{N}\}$ gives the assertion.

(iv) If $a,b$ are real and $a\le b+\varepsilon$ for every real $\varepsilon>0$, then $a\le b$; for otherwise $b<a$, and $\varepsilon=(a-b)/2$ is positive with $a\le b+(a-b)/2$, that is $(a-b)/2\le0$, contradicting $b<a$.

Step 1. Assume that $\bigl(\int_X g\,d\mu_n\bigr)_{n\in\mathbb{N}}$ converges to $\int_X g\,d\mu$ for every bounded Lipschitz $g:X\to\mathbb{R}$. Then for every bounded lower semicontinuous $f:X\to\mathbb{R}$,

$$\int_X f\,d\mu\le\liminf_n\int_X f\,d\mu_n .$$

Let $M\ge0$ be real with $|f(x)|\le M$ for every $x\in X$, and let $h$ be the pointwise sum of $f$ and the constant function with value $M$, so that $0\le h(x)\le 2M$ for every $x\in X$. The constant function with value $M$ is continuous on $X$, hence lower semicontinuous by claim 2 of Semicontinuity Under Negation and Characterization of Continuity, so $h$ is lower semicontinuous on $X$ by claim 3 of Sums and Nonnegative Multiples of Semicontinuous Functions.

Apply Bounded Lower Semicontinuous Functions are Increasing Limits of Lipschitz Functions to $h$ with the bound $2M$: there are functions $h_k:X\to\mathbb{R}$, $k\in\mathbb{N}$, each Lipschitz and continuous on $X$ (claim 2 there), with $0\le h_k(x)\le h(x)\le 2M$ (claim 1), $h_k(x)\le h_{k+1}(x)$ (claim 3), and $h(x)$ the least upper bound of $\{h_k(x):k\in\mathbb{N}\}$ (claim 4), for every $x\in X$. In particular each $h_k$ is bounded and Lipschitz.

Fix $k\in\mathbb{N}$. For every $n$, monotonicity of the integral (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) gives $\int_X h_k\,d\mu_n\le\int_X h\,d\mu_n$. By the assumption of Step 1, $\bigl(\int_X h_k\,d\mu_n\bigr)_{n}$ converges to $\int_X h_k\,d\mu$, so for a real $\varepsilon>0$ there is $N\in\mathbb{N}$ with $\int_X h_k\,d\mu-\varepsilon<\int_X h_k\,d\mu_n$ for every $n\ge N$, and therefore $\int_X h_k\,d\mu-\varepsilon\le\int_X h\,d\mu_n$ for every $n\ge N$. Step 0(i) gives $\int_X h_k\,d\mu-\varepsilon\le\liminf_n\int_X h\,d\mu_n$, and Step 0(iv), applied with $a=\int_X h_k\,d\mu$ and $b=\liminf_n\int_X h\,d\mu_n$, gives

$$\int_X h_k\,d\mu\le\liminf_n\int_X h\,d\mu_n .$$

Each $h_k$ and $h$ is nonnegative, real valued and measurable, hence measurable as a $[0,\infty]$-valued function in the sense of Lebesgue Integral of a Nonnegative Measurable Function, since for every real $a$ the set $\{x\in X:h(x)>a\}$ is the preimage of $(a,\infty)$ and so lies in $\mathcal{B}(X)$; and by claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space the integrals of $h_k$ and of $h$ as nonnegative measurable functions are real and agree with their integrals as integrable functions. The sequence $(h_k)_{k\in\mathbb{N}}$ is nondecreasing with pointwise least upper bound $h$, so Monotone Convergence Theorem shows that $\int_X h\,d\mu$ is the least upper bound of $\{\int_X h_k\,d\mu:k\in\mathbb{N}\}$. By the previous display, $\liminf_n\int_X h\,d\mu_n$ is an upper bound of that set, so

$$\int_X h\,d\mu\le\liminf_n\int_X h\,d\mu_n .$$

Finally, let $\nu$ be any of $\mu$ and the $\mu_n$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral and claim 6(a) of Borel Measurability and Bounded Integration on a Metric Space, $\int_X h\,d\nu=\int_X f\,d\nu+M\,\nu(X)=\int_X f\,d\nu+M$, since $\nu(X)=1$. Step 0(iii) therefore turns the last display into

$$\int_X f\,d\mu+M\le\Bigl(\liminf_n\int_X f\,d\mu_n\Bigr)+M ,$$

and subtracting $M$ proves Step 1.

Claim 1. Assume the hypothesis of claim 1 and let $f:X\to\mathbb{R}$ be bounded continuous, with $|f(x)|\le M$ for every $x$. By claim 2 of Semicontinuity Under Negation and Characterization of Continuity, $f$ is both upper and lower semicontinuous on $X$, and by claim 1 of that lemma the pointwise negation $-f$ is lower semicontinuous on $X$; it satisfies $|-f(x)|\le M$ as well. Write $b_n=\int_X f\,d\mu_n$ and $b=\int_X f\,d\mu$; by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $\int_X(-f)\,d\mu_n=-b_n$ and $\int_X(-f)\,d\mu=-b$. Step 1, applied to $f$ and to $-f$, gives

$$b\le\liminf_n b_n,\qquad -b\le\liminf_n(-b_n).$$

Let $\varepsilon>0$ be real. By claim 3 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence applied to $(b_n)$ there is $N_1\in\mathbb{N}$ with $\liminf_n b_n-\varepsilon<b_m$ for every $m\ge N_1$, hence $b-\varepsilon\le\liminf_n b_n-\varepsilon<b_m$. Applied to $(-b_n)$ there is $N_2\in\mathbb{N}$ with $\liminf_n(-b_n)-\varepsilon<-b_m$ for every $m\ge N_2$, hence $-b-\varepsilon\le\liminf_n(-b_n)-\varepsilon<-b_m$, that is $b_m<b+\varepsilon$. For every $m$ with $N_1\le m$ and $N_2\le m$ we therefore have $b-\varepsilon<b_m<b+\varepsilon$, so $|b_m-b|<\varepsilon$ by claim 9 of Properties of the Absolute Value in an Ordered Field. Hence $(b_n)$ converges to $b$. As $f$ was an arbitrary bounded continuous function, $(\mu_n)$ converges weakly to $\mu$ in the sense of Weak Convergence of Finite Borel Measures on a Metric Space.

Claim 2. Assume $(\mu_n)$ converges weakly to $\mu$. Every bounded Lipschitz $g:X\to\mathbb{R}$ is continuous on $X$ by A Lipschitz Map is Uniformly Continuous, hence bounded continuous, so $\bigl(\int_X g\,d\mu_n\bigr)$ converges to $\int_X g\,d\mu$. Thus the assumption of Step 1 is satisfied, and Step 1 gives the asserted inequality for every bounded lower semicontinuous $f$. The boundedness of $\bigl(\int_X f\,d\mu_n\bigr)$ was noted at the outset.

Claim 3. Let $U\in\mathcal{T}_d$ and let $\mathbf{1}_U$ be its indicator function, so that $0\le\mathbf{1}_U(x)\le1$ for every $x$ and $\mathbf{1}_U$ is bounded. It is lower semicontinuous on $X$: let $x\in X$ and let $\varepsilon>0$ be real. If $x\in U$, then by Open Subset of a Metric Space there is a real $r>0$ with $B_d(x,r)\subseteq U$; every $y\in X$ with $d(x,y)<r$ then lies in $U$, so $\mathbf{1}_U(y)=1>1-\varepsilon=\mathbf{1}_U(x)-\varepsilon$. If $x\notin U$, then $\mathbf{1}_U(x)=0$ and every $y\in X$ satisfies $\mathbf{1}_U(y)\ge0>-\varepsilon=\mathbf{1}_U(x)-\varepsilon$, so any positive $\delta$ serves.

For every probability measure $\nu$ on $(X,\mathcal{B}(X))$, the function $\mathbf{1}_U$ is a nonnegative simple function whose integral in the sense of Simple Function and Its Integral is $\nu(U)$; by the agreement of the two notions of integral for nonnegative simple functions recorded in Lebesgue Integral of a Nonnegative Measurable Function together with claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space, $\int_X\mathbf{1}_U\,d\nu=\nu(U)$. Claim 2 applied to $f=\mathbf{1}_U$ therefore gives $\mu(U)\le\liminf_n\mu_n(U)$.

Claim 4. Let $F$ be closed in $(X,\mathcal{T}_d)$ and put $U=X\setminus F$, which is open by Closed Subset of a Topological Space. The sequences $(\mu_n(F))$ and $(\mu_n(U))$ take values in $[0,1]$ by claim 2 of Basic Properties of a Measure, so they are bounded. Since $\mu$ and the $\mu_n$ are probability measures, hence finite, claim 3 of Basic Properties of a Measure gives $\mu_n(U)=1-\mu_n(F)$ for every $n$ and $\mu(U)=1-\mu(F)$.

Let $\varepsilon>0$ be real. By claim 3 we have $\mu(U)\le\liminf_n\mu_n(U)$, and by claim 3 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence there is $N\in\mathbb{N}$ with $\liminf_n\mu_n(U)-\varepsilon<\mu_m(U)$ for every $m\ge N$. Combining, $\mu(U)-\varepsilon<\mu_m(U)$ for every $m\ge N$, that is $1-\mu(F)-\varepsilon<1-\mu_m(F)$, that is $\mu_m(F)<\mu(F)+\varepsilon$, for every $m\ge N$. Step 0(ii) gives $\limsup_n\mu_n(F)\le\mu(F)+\varepsilon$, and Step 0(iv) then gives $\limsup_n\mu_n(F)\le\mu(F)$.