Suppose $X$ and $Y$ are Polish spaces. Suppose $Z \subset X$ is a borelian subset of $X$. Let $\varphi : Z \to Y$ continuos and $(R_n)_{n \geq 1}$ probabilities on $X$ such that for all $n \geq 1$ $R_n(Z)=1$. Let us define $Q_n$ the image probability of $R_n$ through $\varphi$ and suppose $R_n \xrightarrow{\mathscr{D}} R$ (i.e. $R_n$ converges in distribution to $R$). If $Q$ is the image probability of $R$ through $\varphi$ can I conclude $Q_n \xrightarrow{\mathscr{D}} Q$?
If yes, why?
I have just observed that if $Z$ is closed then is a Polish space and thus we can use reduce to study the case $\varphi : X \to Y$. It is enough to prove $R(A) \leq \liminf R_n(A)$ for any open set $A$.
I do not know hot to approach the case $\varphi: Z \subset X \to Y$.