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?

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