Narrow equals wide topology on probability measures: reference?

Question

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.)

Answer 1

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.