Think of a LCH space $X.$ Consider the spaces $C_{0}(X)$ of continuous functions "vanishing at infinity" and the space $BC(X)$ of bounded continuous functions. Consider as well the space of Radon (Borel regular) measures $M(X).$

What follows is an incorrect reasoning which gets me to an absurd situation. However, I fail to see the mistake(s) in it, and that's where I would like some help!

By the Riesz's representation theorem, we have that the topological dual of $C_{0}(X)$ is isometrically isomorphic to the space of finite Radon measures. Hence, we have in $M(X)$ the natural weak* star topology: We say that $\mu_{n}$ converges weak* to $\mu$ if

$$\int _{X} \psi \ d\mu_{n} \rightarrow \int _{X} \psi \ d\mu, \ \forall \psi \in C_{0}(X).$$

On the other hand, we say that $\mu_{n}$ converges in the narrow topology to $\mu$ if

$$\int _{X} \phi \ d\mu_{n} \rightarrow \int _{X} \phi \ d\mu, \ \forall \phi \in BC(X). $$

Prokhorov's theorem gives a characterization of sequential compactness in $M(X)$ with the narrow topology (actually, of compactness, since it is metrizable). This theorems is quite technical and involves the notion of tightness, which is a necessary condition for compactness.

However, if we think of $M(X)$ with the weak* topology it inherits from being a dual space, compactness is very easy to characterize thanks to the Banach-Alaoglu theorem.

Furthermore, it is a not-too-hard exercise to show the equivalence of the following two propositions:

i) $\mu_{n}$ converges narrowly to $\mu$

ii) $\mu_{n}$ converges weakly* to $\mu$ and $\mu_{n}(X) \rightarrow \mu(X).$

And here is where I got confused: Consider a collection of probability measures, $\{ \mu_{n} \}_{n \in \mathbb{N}}.$ Clearly, this family is bounded in the dual norm. Therefore, by the Banach-Alouglu theorem, it must contain a weakly* convergente subsequence (which I do not relabel) to a measure $\mu$. This subsequence trivilly satisfies $\mu_{n}(X)=1 \rightarrow 1=\mu(X).$ By the remark above, $\mu_{n}$ must converge narrowly to $\mu...$ But the tightness condition has not appeared anywhere!

What is my mistake? What am I doing wrong? Also, if this were so easy, Prokhorov's theorem would be meaningless (of course that's not the case!)

Thank you for your help


The mistake is in

This subsequence trivially satisfies $\mu_n(X) = 1 \to 1 = \mu(X)$.

Consider a sequence $(x_n)$ of points in $X$ converging to $\infty$ (that is, for every compact subset $K$ of $X$ the set $\{n \in \mathbb{N} ; x_n \in K\}$ is finite), and let $\mu_n$ be the point mass in $x_n$ (or $\mu_n = \delta_{x_n}$).

Then you have $\mu_n \to 0$ in the weak$^{\ast}$ topology since $f(x_n) \to 0$ for every $f \in C_0(X)$, and $0$ clearly isn't a probability measure.

The tightness ensures that such things cannot happen. In the example, the mass "escapes to infinity", which tightness prohibits.

  • $\begingroup$ Yes... I realized short after posting that that step was unjustified. Sorry about that, it was pretty obvious, I just got confused... Anyway, thanks for the nice simple counterexample $\endgroup$ – Qwertuy Oct 14 '15 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.