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The following page says that the narrow (sometimes "weak") topology (induced by bounded continuous functions) equals the wide [or weak* topology] (induced by compactly-supported continuous functions) the set $P(X)$ of Radon probability measures on X, provided that X is a Hausdorff space, or I think that X is required to be locally compact Hausdorff.

https://encyclopediaofmath.org/wiki/Convergence_of_measures

Could somebody provide an exact reference to this? I am also interested in exact references on similar results.

Moreover, I'd like to know [Edit:]
(A) if the result is correct (and contained or clearly implied by the some reference); i.e., the wide topology on $P(X)$ equals the narrow topology on $P(X)$; or
(B) if the reference only says that the two topologies have the same convergent sequences; i.e., if $\{\mu_n\}_{n \geq 1}\subset P(X)$, $\mu\in P(X)$, and $$\lim_{n\rightarrow\infty}\int_X f(x) \mu_n(dx) = \int_X f(x) \mu(dx), \tag{1}\label{eq1}$$ for all compactly-supported continuous functions $f:X\to\mathbb K$ (where $\mathbb K$ is $\mathbb R$ or $\mathbb C$), then \eqref{eq1} holds for all bounded continuous functions $f:X\to\mathbb K$.

Of course, I'd like to know the exact conditions required on $X$ and $P(X)$ in the reference. My belief is that $X$ is required to be "any locally compact Hausdorff space" and $P(X)$ is required to be Radon but not necessarily more; i.e., "$P(X)$ = all Radon probability measures" or equivalently, all regular Borel probability measures (see below).

Background information:
Rudin: RCA, Theorem 6.19 says that if $X$ is a locally compact Hausdorff space, then the dual of $C_0(X)$ (hence of $C_c(X)$ too) is exactly the space of regular measures, so then the word "weak*" for the wide topology is justified.

Indeed, for locally compact Hausdorff spaces, Radon measures (i.e., inner regular or tight measures) are the same as regular measures, by https://encyclopediaofmath.org/wiki/Radon_measure (A sufficient criterion is Rudin: 2.18.)

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  • $\begingroup$ For your reference, a generalization of this result can be found here. $\endgroup$
    – Analyst
    Nov 8, 2022 at 22:05
  • $\begingroup$ Can you please explain why the statement in the parenthesis in "the dual of $C_0(X)$ (hence of $C_c(X)$ too) is exactly the space of regular measures"? $\endgroup$ Jul 26, 2023 at 17:39
  • $\begingroup$ Theorem: Let $B$ be the closure of the metric TVS $A$. Then $A^*=B^*$. Proof: Clearly, $B^*\subset A^*$. (If $f,g\in B^*$ and $f=g$ on $A$, then $f=g$ on $B$, by continuity.) Let $f\in A^*$. By proofwiki.org/wiki/… some continuous $F:B\to\mathbb{K}$ extends $f$. By continuity, $F$ is linear, QED. $\endgroup$ Jul 29, 2023 at 9:23
  • $\begingroup$ Theorem: If $A,B$ are TVSs (by which I mean Hausdorff TVS) and $A$ is dense in $B$, and $Z$ is a complete Hausdorff TVS, then every continuous linear $f:A\to Z$ has a unique continuous linear extension $B:\to Z$, by the corollary here (let $C$ be the completion of $B$): en.wikipedia.org/wiki/… $\endgroup$ Jul 29, 2023 at 9:26
  • $\begingroup$ $C_0(X)$ is the completion of $C_c(X)$, by Theorem 3.17 (Rudin: RCA), if $X$ is a locally compact Hausdoff space. In some other books, that holds by the definition of $C_0(X)$. In those cases, $C_0(X)^* = C_c(X)^*$ (when we identify every $f:C_0(X)\to\mathbb{K}$ by its restriction to $C_c(X)$). Here $\K$ is your scalar field ($\mathbb{R}$ or $\mathbb{C}$). I hope that this is what you asked, @ViktorStein. $\endgroup$ Jul 29, 2023 at 9:31

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In your summary of the link you provided, you omitted the most important condition: $X$ has to be compact for weak convergence to be characterized by compactly supported continuous functions. Indeed, they give the canonical example:

However, if $X$ is not compact, the compactness of the wide topology fails: as an example take the sequence of Dirac masses $\delta_n$ on $\mathbb{R}$, where $n\in\mathbb{N}$.

What is true is this: if $\{\mu_n\}_{n \geq 1}$ is a sequence of probability measures such that, for any compactly supported continuous function $f$, $$\lim_{n\rightarrow\infty}\int_X f(x) \mu_n(dx) = \int_X f(x) \mu(dx),$$ and in addition $\mu$ is a probability measure on $X$, then $\mu_n$ converges weakly (in distribution) to $\mu$.

This result is Theorem 7.7 on p. 95 of Khoshnevisan's Probability. It is a special case of the tightness criterion on the page you linked. The standard reference is Billingsley's Convergence of probability measures.

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  • $\begingroup$ Thank you. It says: for $X=\mathbb R^k$ (usual Borel sets) the "(B)" in my question holds, but it says nothing about (A). Is (A) even true? (Weak convergence "=>" is defined in Definition 7.1 on pp. 91-92 and it means convergence w.r.t. the weak or "narrow" topology.) books.google.fi/books?id=BgYPCgAAQBAJ&pg=PA95 $\endgroup$ Sep 1, 2020 at 8:45
  • $\begingroup$ The result you provided is valuable to me, though it would be even better to have a more general $X$ and/or to know if also (A) is true. AFAIS, your counter-example only shows that convergence to a measure (namely $\mu=0$) does not imply that $\mu\in P(X)$. But Theorem 7.7 does not require compactness. $\endgroup$ Sep 1, 2020 at 9:23

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