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).
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> **Conclusion:** Then the Borel $\sigma$-algebras generated by $T\_N$ and $T\_{WC}$ are the same.
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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 $X$.
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> **Conclusion:** Then $A$ is equal to the Borel $\sigma$-algebras generated by $T\_N$.
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When do Borel $\sigma$-algebras generated by the total variation norm and the weak* topology coincide?
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.


