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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

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    $\begingroup$ How can I show ii) implies i)? $\endgroup$
    – Focus
    Commented Jul 18, 2019 at 7:58
  • $\begingroup$ Could you explain how to show (ii) implies (i)? I think it's not trivial at all. $\endgroup$
    – Analyst
    Commented Nov 1, 2022 at 21:12

1 Answer 1

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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.

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    $\begingroup$ 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
    Commented Oct 14, 2015 at 13:21
  • $\begingroup$ I've just posted a related question here. Could you please have a look at it? $\endgroup$
    – Analyst
    Commented Nov 1, 2022 at 21:13

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