Prokhorov's Theorem-Prove if tight subsubsequence, then tight sequence. Let $P_n$ be a sequence of Borel probability measures on $\mathbb{R}$ has a subsequence $\{P_n\}_k$ has a further subsequence that is tight. Show that $P_n$ is tight.
Clearly, this is Prokorov's Theorem. I have that the subsubsequence $\{\{P_n\}_k\}_l$ to $P$ weakly. Does this further imply that $\{P_n\}_k$ converges weakly to $P$ which would make $P_n$ tight? Or am I assuming too much here?
 A: It seems that there is a mistake in the statement: this should be 

Let $P_n$ be a sequence of Borel probability measures on $\mathbb{R}$ such that each subsequence $\{P_{n_k}\}_k$ has a further subsequence that is tight.

Otherwise, if $P_{2n}=P$ and $\left(P_{2n+1}\right)_{n\geqslant 1}$ is not tight, we have a counter-example. 
There is no need to use Prokhorov's theorem. We can argue as follows: if the sequence $(P_{n})_{n\geqslant 1}$ is not tight, then there is $\delta_0\gt 0$ such that for each $j$, we can find infinitely many integers $i$ such that $P_{i}\left(\mathbb R\setminus \left[-j,j\right]\right)\gt \delta_0$.
Otherwise, for all $\delta$, there exists $j$ such that the set of integers $i$ such that $P_{i}\left(\mathbb R\setminus \left[-j,j\right]\right)\gt \delta$ is finite. Using tightness of a finite family of probability measures, we would obtain tightness of $(P_{n})_{n\geqslant 1}$
Therefore, we can construct inductively an increasing sequence of integers $(n_j)_{j\geqslant 1}$ such that $$\forall j\geqslant 1, \quad P_{n_j}\left(\mathbb R\setminus \left[-j,j\right]\right)\gt \delta_0.$$
The sequence $\left(P_{n_j}\right)_{j\geqslant 1}$ does not admit any tight subsequence.
