When do Borel $\sigma$-algebras generated by the total variation norm and the weak* topology coincide?

I am almost certain that I read somewhere that the following is true, but I cannot seem to locate the reference. I would be most appreciative if someone could point me to a reference. The result was in one of the following forms.

Premise: Let $X$ be a Polish space. Let $S$ be the vector space of totally finite signed Borel measures on $X$. Let $T_N$ be the topology induced by the total variation norm on $S$. Let $T_{WC}$ be the topology of weak convergence of measures on $S$ (sometimes called the weak* topology).

Conclusion: Then the Borel $\sigma$-algebras generated by $T_N$ and $T_{WC}$ are the same.

Other form (which would imply the above):

Premise: Let $X$ be a measurable space with a countably generated $\sigma$-algebra. Let $S$ be the vector space of totally finite signed measures on $X$. Let $T_N$ be the topology induced by the total variation norm on $S$. Let $A$ be the $\sigma$-algebra on $S$ generated by maps of the form $\mu \mapsto \int_X f d\mu$ where $f\colon X \to \mathbb{R}$ varies over bounded measurable real-valued functions on $X$.

Conclusion: Then $A$ is equal to the Borel $\sigma$-algebras generated by $T_N$.

No, it isn't true.

Let $E$ be a non-Borel set in $X$ and consider the set of Dirac measures $E' = \{ \delta_x : x \in E\} \subset S$. Note that for each Dirac measure $\mu$, the total-variation ball of radius 1/2 about $\mu$ contains no other Dirac measure. Hence $E'$ is closed in $T_N$, hence Borel in $T_N$.

On the other hand, it is well known that the map $F : X \to (S, T_{WC})$ defined by $x \mapsto \delta_x$ is continuous (in fact, a homeomorphism onto its image). In particular, it is Borel measurable. Since $E = F^{-1}(E')$ is not Borel in $X$, we conclude that $E'$ is not Borel in $T_{WC}$.

A statement that is true is that the total-variation norm is a Borel function on $(S, T_{WC})$; indeed, it is lower semi-continuous. This is an immediate consequence of the identity $$\|\mu\| = \sup \left\{ \int \varphi \,d\mu : \varphi \in C_b(X), \|\varphi\|_\infty \le 1\right\}$$ which follows from the fact that each finite $\mu$ on a Polish space is regular. So every total-variation ball is $\sigma(T_{WC})$ measurable. But since $T_N$ is not second countable, we can't conclude that the open balls generate $\sigma(T_N)$, and indeed they do not.

• What if one restricts to a class of absolutely continuous probability measures (or a class of probability measures which have at most finitely many atoms at fixed points, say $x_1, \ldots, x_k$? In that case do the two $\sigma$-algebras coincide? – Jack London Apr 21 at 16:40
• @JackLondon: If all your measures are supported on a fixed finite set, then your space of measures is finite dimensional and the two topologies are the same, hence their Borel $\sigma$-algebras are also the same. – Nate Eldredge Apr 21 at 16:45
• @JackLondon: If all your measures are absolutely continuous to some fixed measure $\mu_0$, then I think it is again true, because now $(S, T_N)$ is isometric to $L^1(\mu_0)$, which is separable and second countable. The argument above shows that every ball is $\sigma(T_{WC})$ measurable, and now $\sigma(T_N)$ is generated by balls. – Nate Eldredge Apr 21 at 16:48
• I've not been sufficiently clear: the space on which measures are defined, say $X$, could be infinite. Fix a finite subset $\{x_1, \ldots, x_m\}\subset X$; then I'm considering the class of all the probability measures of the form $\mu=\alpha_0 \mu_{AC} + \sum_{i=1}^m \alpha_i\delta_{x_i}$ where $\mu_{AC}$ is a probability measure on $X$ absolutely continuous w.r.t. a fixed measure $\nu$ (e.g. Lebesgue if $X \subset \mathbb{R}^d$) and $\alpha_0+\alpha_1+\ldots+\alpha_m=1$. – Jack London Apr 21 at 16:51
• For such a class, could the two $\sigma$-algebra coincide? – Jack London Apr 21 at 16:53