Let $\mathbb{X}$ be a complete metric space, $U(\mathbb{X})$ the space of bounded and continuous functions in $\mathbb{X}$ and $\mathcal{L}\big(U(\mathbb{X})\big)$ the space of all linear functionals $L:U(\mathbb{X})\to\mathbb{R}$.
By definition, the weak topology of $U(\mathbb{X})$ is the smallest topology of $U(\mathbb{X})$ $\Big($"smallest" with respect to lower order of inclusion "$\subset$" in $\{ \tau : \tau \mbox{ is topology of } U(\mathbb{X})\}$ $\Big)$ that makes continuous all linear functionals $L:U(\mathbb{X})\to\mathbb{R} $ of $\mathcal{L}\big(U(\mathbb{X})\big)$.
A linear functional $L\in\mathcal{L}\big(U(\mathbb{X})\big) $ is called positive if $f\geq 0$ implies $L(f)\geq 0$, $\forall f\in U(\mathbb{X})$. Let $\mathcal{L}_{\geq 0}\big( U(\mathbb{X})\big)$ the subspace of all linear functionals positives of $\mathcal{L}\big( U(\mathbb{X}) \big)$.
The Riesz Markov Theorem tells us that the space of positive linear functional $\mathcal{L}_{\geq 0}\big( U(\mathbb{X})\big)$ and $\mathcal{M}(\mathbb{X})$ the space of measures with sign $\mu$ on Borel subsets of $\mathbb{X}$ are isomorphic. So it makes sense to speak of the weak topology of $\mathcal{M}(\mathbb{X})$ which is the topology induced by the isomorphism.
But several authors of books on probability in metric spaces ( see for exemple Parthasarathy p. 40 ) define the weak topology in $\mathcal{M}(\mathbb{X})$ as that generated by the following system of neighborhoods:
$ V_\mu \big( f_1,\dots,f_n,\epsilon_1,\dots\epsilon_n\big)=\bigg\{ \nu\in \mathcal{M}(\mathbb{X}) : \bigg| \int_{\mathbb{X}} f_i d\mu -\int_{\mathbb{X}}f_i d\nu \;\bigg|<\epsilon_i \bigg\} $ whit $ f_1,\dots,f_n\in U(\mathbb{X})$.
Question: as we prove that these two topologies in space $\mathcal{M}(\mathbb{X})$ are really equals?