Weak convergence of pushfoward measures

Question

Given $P_n$ and $P$ probability measures over $(X, d_X)$, show that, if $(Y,d_y)$ is another metric space and $\psi: X\rightarrow Y$ is continous, then, "$P_n\Rightarrow P" \Rightarrow \psi_\text{#}P_n\Rightarrow \psi_\text{#}P$.

By $\psi_\text{#}P_n$ I mean the pushfoward measure. It seems to be true, but using the definition of weak convergence, it is clear that what we must show, but, unfortunately, I am struggling with it. Can someone give me a hand?

Answer 1

Recall that by the definition of the push-forward measure we have for any $f$ that \begin{equation} \int f(x)d[\psi_\# P_n](x) = \int f(\psi(x))dP_n(x). \end{equation}

Now let $f$ be a continuous and bounded function with respect to $(Y,d_Y)$. Note that a composition of continuous functions is also continuous and therefore $f \circ \psi$ is continuous. Also if $f$ is bounded then of course $f \circ \psi$ is also bounded. In particular weak convergence of $P_n$ to $P$ implies \begin{equation} \lim_{n \rightarrow\infty}\int f(\psi(x))dP_n(x) =\int f(\psi(x)) dP(x). \end{equation}

Putting both facts together we have \begin{align} \lim_{n \rightarrow \infty}\int f(x)d[\psi_\# P_n](x) &= \lim_{n \rightarrow\infty}\int f(\psi(x))dP_n(x) & (\text{Def. of Push-forward}) \\ &=\int f(\psi(x)) dP(x) & (\text{Weak convergnce}) \\ &= \int f(x) d[\psi_\#P](x). & (\text{Def. of Push-forward}) \end{align} where the last equality uses the first fact but from right-to-left. This shows that for all bounded continuous functions we indeed have \begin{equation} \lim_{n \rightarrow \infty}\int f(x)d[\psi_\# P_n](x) = \int f(x) d[\psi_\#P](x) \end{equation} and therefore $\psi_{\#}P_n$ converges weakly to $\psi_\#P$.

Answer 2

Recall, from Portmaneau theorem, that, given $B \subset X$ open, then $$\liminf_{n} P_n(B)>P(B)$. Therefore,

\begin{equation} \liminf_{n} \psi_{\text{#}}P_n(A) = \liminf_n P_n(\psi^{-1}(A))\\ \geq P(\psi^{-1}(A))\\ =\psi_{\text{#}}P(A) \end{equation}

Since $\psi_{-1}(A)$ is open, because $A$ is open. Therefore, $\liminf_{n} \psi_{\text{#}}P_n(A) \geq P(\psi^{-1}(A))$, and, from Portmanteau, $\psi_{\text{#}}P_n \Rightarrow \psi_{\text{#}}P$