I am currently reading the book Convergence of Probability Measures by Patrick Billingsley, and I came across the following theorem:

Theorem. Let $(S,\rho)$ be a metric space, and $B(S)$ be the Borel $\sigma$-algebra (generated by the metric topology induced by $\rho$). Let $\{P_n:n\in\mathbb N\}$ and $P$ be probability measures on $(S,B(S))$. Then, $P_n$ converges weakly to $P$ if and only if every subsequence $\{P_{n_k}:k\in\mathbb N\}$ contains a further subsequence $\{P_{n_{k(i)}}:i\in\mathbb N\}$ which converges weakly to $P$.

Now, I don't have any problem with the proof of this, but I question wether this could be useful in demonstrating that a sequence of probability measures converge weakly (the book does not provide any example).

In other words, are there examples of sequences of probability measures $\{P_n\}$ such that verifying that each subsequence has a weakly convergent subsequence is any easier than verifying weak convergence directly? Or maybe this can be useful in demonstrating that a sequence does not converge weakly by finding a subsequence with no convergent subsequence?


1 Answer 1


This property is not specific to weak convergence. It indeed workw replacing the term "weak convergence" with "convergence in $(X,\mathcal T)$", where $(X,\mathcal T)$ is a topological space.

However, in the concept of probability measures, we can relate it with the concept of tightness.

It helps for example in the concept of measures in $C[0,1]$.

Proposition. Let $(\mu_n)_n$ be a sequence of Borel probability measures on $C[0,1]$ for which the finite dimensional distributions converge to those of a measure $\mu$. If $\{\mu_n,n\geqslant 1\}$ is tight, then the sequence $(\mu_n)_{n\geqslant 1}$ converges in distribution to $\mu$.

  • $\begingroup$ I see. So its usefulness basically comes from the fact that it helps in the demonstrations of Prokhorov's Theorem and the proposition you mentioned? $\endgroup$
    – user75206
    Apr 10, 2014 at 15:39
  • 1
    $\begingroup$ No, Prokhorov's theorem is proven differently. It helps to show weak convergence of a sequence showing tightness, and the uniqueness of a potential limit. $\endgroup$ Apr 10, 2014 at 20:29

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