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?

  • $\begingroup$ Yes. Actually, $\tau=\tau_d$. $\endgroup$ Feb 2, 2021 at 20:01
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    $\begingroup$ Are you sure? I think $\tau$ is not metrisable... $\endgroup$
    – Bremen000
    Feb 2, 2021 at 20:09
  • $\begingroup$ You are right, I misread what you wrote and thought you were writing about positive measures. $\endgroup$ Feb 2, 2021 at 20:10
  • $\begingroup$ 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... $\endgroup$
    – Bremen000
    Feb 2, 2021 at 20:16
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    $\begingroup$ 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... $\endgroup$
    – Bremen000
    Feb 3, 2021 at 12:49

1 Answer 1


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

  • $\begingroup$ 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... $\endgroup$
    – Bremen000
    Feb 3, 2021 at 11:13
  • $\begingroup$ 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 $\sigma(\tau)\subseteq \sigma(\tau_d)$. $\endgroup$ Feb 3, 2021 at 11:47
  • $\begingroup$ 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? $\endgroup$
    – Bremen000
    Feb 3, 2021 at 11:53
  • $\begingroup$ 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. $\endgroup$ Feb 3, 2021 at 11:54
  • $\begingroup$ Oh thanks god: I started questioning everything! $\endgroup$
    – Bremen000
    Feb 3, 2021 at 11:55

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