# 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?

• Yes. Actually, $\tau=\tau_d$. Feb 2, 2021 at 20:01
• Are you sure? I think $\tau$ is not metrisable... Feb 2, 2021 at 20:09
• You are right, I misread what you wrote and thought you were writing about positive measures. Feb 2, 2021 at 20:10
• Yes, in case I restrict $\tau$ to positive measures this is true. I thought to the following argument but I am not sure: if $B \in \sigma(\tau_d)$ I can consider the sets $B_n := B \cap \{ |\mu| \le n \}$. I think that $\tau |_{B_n} = \tau_d |_{B_n}$ (I am not completely sure) so that $B_n \in \sigma(\tau)$. However $B= \cup_n B_n$, so that $B \in \sigma(\tau)$ and then $\sigma(\tau_d) \subset \sigma(\tau)$. Let me know if it convinces you... Feb 2, 2021 at 20:16
• Actually I can define $\{f_k\}_{k \ge 1} = \{ 1 \} \cup \left (\cup_n \cup_m \cup_j f_{m,n,j} \right )$ where $f_{m,n,j}$ is a smooth function identically one on $B_{1/m}(x_n)$ and identically $0$ outside $B_{1/m +1/j}(x_n)$, where $\{x_n\}_{n \ge 1} = \mathbb{Q}^d$. In this way all the functions are already nonnegative and uniformly bounded. I didn't specify it because I thought it was not important... Feb 3, 2021 at 12:49

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 $$(\mathcal M,\tau_d)\to\mathcal M$$.

• The argument above allows me to conclude that $\sigma(\tau) \subset \sigma(\tau_d)$. Is it not already obvious from $\tau \subset \tau_d$? I need the opposite inclusion... Feb 3, 2021 at 11:13
• The obvious direction is $\sigma(\tau_d)\subseteq \sigma(\tau)$ because the metrizable topology is contained in the weak topology. What I have written is the non-obvious direction $\sigma(\tau)\subseteq \sigma(\tau_d)$. Feb 3, 2021 at 11:47
• I’m sorry, I am confused: if $E$ is $\tau$-closed, then $E$ is sequentially $\tau$-closed. Then it is sequentially $\tau_d$-closed i.e. it is $\tau_d$-closed hence $\tau\subset \tau_d$. Where is my mistake? Feb 3, 2021 at 11:53
• Ups! I have just understood that your metric $d_p$ does not generate the weakest topology in which all integrals $\int f_kd\mu$ are continuous. So, this changes the situation. Let me think a bit more. Feb 3, 2021 at 11:54
• Oh thanks god: I started questioning everything! Feb 3, 2021 at 11:55