# 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?

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.