Weak convergence vs weak* convergence of a measure

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 $$C_b(X)$$?

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 $$\int fd\mu_{n}=1\neq\int fd\mu$$.