Dose weak convergence imply tight? $X$ is a separable metric space, $\{P_n\}_{n=1}^{\infty}$, $P$ are probability measures on $X$, and $P_n$ converges weakly to $P$, can we conclude that $\{P_n\}_{n=1}^{\infty}$ is tight?
I know if $X$ is also complete, then it is right by Prokhorov theorem.
 A: Your question is related to the concepot of a universally measurable metric space. All definitions and Theorems below are taken from the book "Real Analysis and Probability" by R.M. Dudley.
A separable metric space $(S,d)$ is universally measurable if for every (Borel) probability measure $P$ on the completion $T$ of $S$, there are $A \subset S \subset B$ with $A,B \subset T$ Borel measurable and with $P(A) = P(B)$, see the Definition in $11.5.
One can then show (cf. Theorem 11.5.1) that a separable metric space $(S,d)$ is universally measurable iff every (Borel) probability measure on $S$ is tight.
Finally (cf. Theorem 11.5.3) a result of Le Cam states that if $(S,d)$ is a metric space and if $P_n$ is tight for every $n$ and $P_n \to P_0$, then $(P_n)_n$ is uniformly tight.
In particular, if $(S,d)$ is separable and universally measurable, then every converging sequence of (Borel) probability measures is uniformly tight.
Hence, your result holds if your space under consideration is universally measurable. One condition when this holds is if $X$ is a Borel subset of its completion.
But if $X$ is not universally measurable, then (by Theorem 11.5.1 mentioned above), there is a law $P$ on $X$ which is not tight. Now, if you set $P_n =P$ for all $n$, then $P_n \to P$, but $(P_n)_n$ is not (uniformly) tight.
Thus, your desired result holds iff $X$ is universally measurable.
