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I am reading Optimal Transport for Applied mathematicians by Filippo Santambrogio and in the first chapter he describes weak convergence and weak-* convergence in terms of a Banach space and its dual. He then goes on to (by an abuse of notation, apparently) extend this concept to signed measures on a separable, locally compact metric space $X$ when he states the Riesz representation theorem by saying:

Let $\mathcal{X} = C_0(X)$ be the space of continuous function on X vanishing at infinity, i.e. $f \in C_0(x) \iff f \in C(X)$, and for every $\epsilon > 0$, there exists a compact subset $K \subset X$ such that $|f| < \epsilon$ on $X$ \ $K$. Let us endow this space with the sup norm since $C_0(X) \subset C_b(X)$ (this last space being the space of bounded continuous functions on $X$). Note that $C_0(X)$ is a Banach space and thaat it is a closed subset of $C_b(X)$. Then every element of $\mathcal{X}'$ (the dual of $\mathcal{X}$) is represented in a unique way as an element of $\mathcal{M}(X)$: for all $\xi \in \mathcal{X}'$ there exists a unique $\lambda \in \mathcal{M}(X)$ such that $\langle \xi, \phi \rangle = \int \phi d \lambda$ for every $\phi \in \mathcal{X}$; moreover, $\mathcal{X}'$ is isomorphic to $\mathcal{M}(X)$ endowed with the norm $||\lambda|| := |\lambda|(X)$.

For signed measures of $\mathcal{M}(X)$, we should call weak-* convergence the convergence in the duality with $C_0(X)$. Yet another interesting notion of convergence is that in duality with $C_b(X)$. We will call it (by abuse of notation) weak convergence and denote it through the symbol $\mu_n \rightharpoonup \mu$ iff for every $\phi \in C_b(X)$ we have $\int \phi d\mu_n \rightarrow \int \phi d \mu$.

My question is as follows: is weak-* convergence (i.e. weak convergence in duality with $C_0(X)$) the same as weak convergence in duality with $C_b(X)$, except for the fact that in the former condition we require $\int \phi d\mu_n \rightarrow \int \phi d \mu$ for all $\phi \in C_0(X)$ as opposed to $C_b(X)$?

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Firstly, we remark that $\mathcal{M}(X)$, as a set, is the collection of all finite signed Radon measures (assuming that we are working with the scalar field $\mathbb{R}$). That is, $\mu\in\mathcal{M}(X)$ if $\mu$ is a finite signed measure, outer-regular in the sense that for each Borel subset $A\subseteq X$, $\mu(A)=\inf\{\mu(U)\mid A\subseteq U\mbox{ and }U\mbox{ is open}\}$, and inner-regular for open set in the sense that for each open set $U\subseteq X$, $\mu(U)=\sup\{\mu(K)\mid K\subseteq U\mbox{ and }K\mbox{ is compact.}\}$. Note that if $X$ is second countable, then every finite signed measure is automatically Radon. (see Folland, Real Analysis, Theorem 7.8).

For your question, the two modes of convergence are not the same. Obviously, given a sequence $(\mu_{n})$ in $\mathcal{M}(X)$ and $\mu\in\mathcal{M}(X)$, $\mu_{n}\rightharpoonup\mu\Rightarrow\mu_{n}\rightarrow\mu$ in weak*-topology. The converse does not hold in general. The following is a counter-example.

Let $X=\mathbb{R}$, equipped with the usual topology. For each $n\in\mathbb{N}$, let $\mu_{n}(A)=\lambda(A\cap[n,n+1])$, where $\lambda$ is the usual Lebesgue measure, then $\mu_{n}\in\mathcal{M}(X)$. Let $\mu=0$. Clearly $\mu_{n}\rightarrow\mu$ in weak*-topology. However, $\mu_{n}\not\rightharpoonup\mu$ because the constant function $f=1$ is an element in $C_{b}(X)$ and $\int fd\mu_{n}=1\neq\int fd\mu$.

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