Borel sigma algebra on measures generated by distance inducing weak convergence and the one generated by weak topology 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?
 A: 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 $(\mathcal M,\tau_d)\to\mathcal M$.
