Let $\mathcal{M}$ be the vector space of Borel finite signed measures on $\mathbb{R}^d$. On $\mathcal{M}$ we can consider the weak topology $\tau$: the coarsest topology on $\mathcal{M}$ s.t. all the maps $\mu \mapsto \int \varphi d\mu$ are continuous on varying of $\varphi \in C\_b(\mathbb{R}^d)$, the continuous and bounded real valued functions on $\mathbb{R}^d$.


Suppose $\{f\_k\}\_{k \ge 1} \subset C\_b(\mathbb{R}^d)$ is a sequence of functions s.t. $\sup\_k \sup\_x |f\_k(x)| \le 1$ and s.t.
$$\mu\_n \overset{\tau}{\to}\mu \text{ iff } \int f\_k d \mu\_n \to \int f\_k d \mu \quad \forall \, k \ge 1.$$


Then we can define the distance $d$ on $\mathcal{M}$ as $$ d(\mu, \nu) = \sum\_k 2^{-k} \left | \int f\_k d \mu - \int f\_k d \nu \right |$$


and we have a topology on $\mathcal{M}$ generated by $d$, call it $\tau\_d$. Of course $\tau \subset \tau\_d$ (but they have the same converging sequences) and then $\sigma(\tau) \subset \sigma(\tau\_d)$, where $\sigma(\mathcal{E})$ denotes the smallest sigma algebra containing $\mathcal{E} \subset 2^{\mathcal{M}}$.


Is it possible to prove also the opposite inclusion i.e. that the Borel sigma algebra generated by those two topologies actually coincide?



Borel sigma algebra on measures generated by distance inducing weak convergence and the one generated by weak topology
The Borel $\sigma$-algebras generated by these two topologies seem to be equal.


The idea of the proof is as follows. Let $\mathcal M\_+$ be the subspace of $\mathcal M$ consisting of measures. It is known that the weak topology on $\mathcal M\_+$ is metrizable and the space $\mathcal M\_+$ is Polish. Consider the subspace $$\mathcal P=\{(\lambda,\mu)\in\mathcal M\_+\times\mathcal M\_+:\lambda\perp\mu\}.$$ The symbol $\lambda\perp\mu$ means that there are disjoint $\sigma$-compact subsets $A,B\subseteq\mathbb R^d$ such that $\lambda(A)=\lambda(\mathbb R^d)$, $\mu(B)=\mu(\mathbb R^d)$ and $\lambda(B)=\mu(A)=0$.
It can be shown that the set $\mathcal P$ is Borel (of type $F\_{\sigma\delta}$) in $\mathcal M\_+\times\mathcal M\_+$.


Now consider the map $$r:\mathcal P\to\mathcal M,\quad r:(\lambda,\mu)\mapsto\lambda-\mu$$and observe that it is continuous and bijective (as each sign-measure uniquely decomposes into its positive and negative parts).
Since $\sup\_{k\in\mathbb N}\|f\_k\|<\infty$, the map $r$ also is also continuous with respect to the topology $\tau\_d$ on $\mathcal M$.


Since the Tychonoff space $\mathcal M$ is a continuous image of the metrizable separable space $\mathcal P$, it has countable network of the topology and hence admits a continuous injective map $\psi:\mathcal M\to \mathbb R^\omega$ to the Polish space $\mathbb R^\omega$.


For any $\tau\_d$-open set $U\subseteq \mathcal M$ the preimage $r^{-1}[U]$ is an open set in $\mathcal P$. By the classical Lusin-Souslin Theorem (15.1 in Kechris' book), the image of any Borel subset of $\mathcal P$ under the injective continuous map $\psi\circ r$ is Borel in the Polish space $\mathbb R^\omega$. In particular, the set $V=\psi\circ r[r^{-1}[U]]$ is Borel in $\mathbb R^\omega$ and hence the set $U=\psi^{-1}[V]$ is Borel in $\mathcal M$. This implies that the Borel $\sigma$-algebra $\sigma(\tau\_d)$ generated by the topology $\tau\_d$ is contained in the Borel $\sigma$-algebra $\sigma(\tau)$ generated by the topology $\tau$. On the other hand, the inclusion $\sigma(\tau)\subseteq \sigma(\tau\_d)$ follows from the metrizability of the topology $\tau\_d$ and the sequential continuity of the indentity map $(\mathcal M,\tau\_d)\to\mathcal M$.


