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, $\tau=\tau_d$. $\endgroup$ – Michael Greinecker Feb 2 at 20:01
  • 1
    $\begingroup$ Are you sure? I think $\tau$ is not metrisable... $\endgroup$ – Bremen000 Feb 2 at 20:09
  • $\begingroup$ You are right, I misread what you wrote and thought you were writing about positive measures. $\endgroup$ – Michael Greinecker Feb 2 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 at 20:16
  • 1
    $\begingroup$ 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 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 $(\mathcal M,\tau_d)\to\mathcal M$.

  • $\begingroup$ 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 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 $\sigma(\tau)\subseteq \sigma(\tau_d)$. $\endgroup$ – Taras Banakh Feb 3 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 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. $\endgroup$ – Taras Banakh Feb 3 at 11:54
  • $\begingroup$ Oh thanks god: I started questioning everything! $\endgroup$ – Bremen000 Feb 3 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.