MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prokhorov theorem provides a useful characterization of relatively compact sets w.r.t. narrow topology (topology induced by narrow convergence) in the space of probability measure.

Notation used throughout:-


$\mathcal{P}(X)$- Space of Borel probability measures on X

$C_b(X)$- Space of continuous and bounded functions on X


Narrow Convergence: A sequence $(\mu_n)\subset\mathcal{P}(X)$ is narrowly convergent to $\mu\in\mathcal{P}(X)$ if $\int_Xfd\mu_n\xrightarrow{n\rightarrow \infty}\int_Xfd\mu$ for every $f\in C_b(X)$.

Tightness: A set $\mathcal{K}\subset\mathcal{P}(X)$ is tight if $\forall \epsilon>0 \ \ , \exists K_\epsilon \text{ compact in $X$ such that } \mu(X\backslash K_\epsilon)\leq \epsilon \ \ \forall \mu\in\mathcal{K}$.

Prokhorov's theorem: If a set $\mathcal{K}\subset \mathcal{P}(X)$ is tight then $\mathcal{K}$ is relatively compact in $\mathcal{P}(X)$.

Question: Does there exists a generalization of Prokhorov's theorem to $\mathcal{M}(X)$, the space of finitely additive signed measures on X? Any references would be welcome.

There is a nice generalization of the Prokhorov's theorem to the space of Borel measures. For proof see Bogachev's Measure Theroy Vol 2 (Thm 8.6.2).

Let $X$ be a complete seperable metric space and $\mathcal{M}$ a family of Borel measures on $X$. Then the following statements are equivalent:

(1) Every sequence $\mu_n\subset\mathcal{M}$ contains a weakly convergent susequence.

(2) The family $\mathcal{M}$ is tight and uniformly bounded in the total variation norm.

share|cite|improve this question
Please don't post on both MSE and MO at the same time.… – Asaf Karagila Aug 29 '13 at 20:25
@UPS So where exactly does the proof of the Prokhorov's theorem fail for finitely additive positive measures with bounded total mass? Also, exactly what statement do you want for signed finitely additive measures? – fedja Sep 6 '13 at 17:09
Since my previous comment was deleted, I repeat it here: how does your reference to the result in Bogachev's result have any relevance to the setting of finitely additive measures (which are more like elements in the dual of $L^\infty({\bf R}^n)$). Like @fedja I am not sure what you actually want. – Yemon Choi Sep 6 '13 at 20:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.