Proof of Weak Convergence of Finite Borel Measures is Preserved by Continuous Maps between Metric Spaces

Each result cited below is universally quantified over the data in its own statement and is applied with the data named at the point of citation.

Write $d_{\mathbb{R}}(s,t)=|s-t|$ for the absolute-value metric on $\mathbb{R}$; by claim 2 of Borel Measurability and Bounded Integration on a Metric Space the Borel $\sigma$-algebra of $(\mathbb{R},d_{\mathbb{R}})$ is the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line. Let $g:Z\to\mathbb{R}$ be bounded, with a bound $M$, and continuous on $Z$. By Weak Convergence of Finite Borel Measures on a Metric Space it suffices to show that $\bigl(\int_{Z}g\,d(T_{\#}\mu_{j})\bigr)_{j\in\mathbb{N}}$ converges to $\int_{Z}g\,d(T_{\#}\mu)$; note that $T_{\#}\mu_{j}(Z)=\mu_{j}(Y)<\infty$ and $T_{\#}\mu(Z)=\mu(Y)<\infty$, as recalled in the statement from claim 1 of Image Measures, Measures with Densities, and Change of Variables, so all these measures are finite Borel measures on $(Z,d_{Z})$.

Step 1 (the composite is a bounded continuous function). Let $u=g\circ T:Y\to\mathbb{R}$. Then $|u(y)|=|g(T(y))|\le M$ for every $y\in Y$, so $u$ is bounded. Let $y\in Y$ and let $\varepsilon$ be a positive real number. Since $g$ is continuous at $T(y)$, Continuous Map Between Metric Spaces gives a positive real $\delta_{1}$ such that every $z\in Z$ with $d_{Z}(T(y),z)<\delta_{1}$ satisfies $d_{\mathbb{R}}(g(z),g(T(y)))<\varepsilon$. Since $T$ is continuous at $y$, the same definition, applied with the positive number $\delta_{1}$, gives a positive real $\delta$ such that every $y'\in Y$ with $d_{Y}(y,y')<\delta$ satisfies $d_{Z}(T(y'),T(y))<\delta_{1}$, hence $d_{Z}(T(y),T(y'))<\delta_{1}$ by the symmetry condition 3 of Metric Space, hence $d_{\mathbb{R}}(u(y'),u(y))<\varepsilon$. These choices are made in the order $\varepsilon$, $\delta_{1}$, $\delta$. Thus $u$ is continuous at every point of $Y$, that is, continuous on $Y$.

Step 2 (measurability and integrability). By claim 3 of Borel Measurability and Bounded Integration on a Metric Space, applied to the continuous maps $g:Z\to\mathbb{R}$ and $u:Y\to\mathbb{R}$ with values in $(\mathbb{R},d_{\mathbb{R}})$, the function $g$ is measurable with respect to $\mathcal{B}(Z)$ and $\mathcal{B}(\mathbb{R})$, and $u$ is measurable with respect to $\mathcal{B}(Y)$ and $\mathcal{B}(\mathbb{R})$. Since $|g|\le M$, claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space shows that $g$ is integrable with respect to every finite Borel measure on $(Z,d_{Z})$, in particular with respect to $T_{\#}\mu_{j}$ for every $j$ and with respect to $T_{\#}\mu$.

Step 3 (change of variables). Let $\nu$ be one of the measures $\mu_{j}$, $j\in\mathbb{N}$, or $\mu$. By claim 2 of Image Measures, Measures with Densities, and Change of Variables, applied to the measure space $(Y,\mathcal{B}(Y),\nu)$, the measurable space $(Z,\mathcal{B}(Z))$, the measurable map $T$ and the measurable function $g$, which is integrable with respect to $T_{\#}\nu$ by Step 2, the function $u=g\circ T$ is integrable with respect to $\nu$ and

$$\int_{Z}g\,d(T_{\#}\nu)=\int_{Y}g\circ T\,d\nu .$$

Step 4 (conclusion). By Step 1, $u$ is a bounded function on $Y$ that is continuous on $Y$, so the hypothesis $\mu_{j}\Rightarrow\mu$ and Weak Convergence of Finite Borel Measures on a Metric Space show that $\bigl(\int_{Y}u\,d\mu_{j}\bigr)_{j\in\mathbb{N}}$ converges to $\int_{Y}u\,d\mu$. By Step 3 these are the numbers $\int_{Z}g\,d(T_{\#}\mu_{j})$ and $\int_{Z}g\,d(T_{\#}\mu)$. As $g$ was an arbitrary bounded function continuous on $Z$, Weak Convergence of Finite Borel Measures on a Metric Space gives $T_{\#}\mu_{j}\Rightarrow T_{\#}\mu$.