Does the $\sigma$-algebra generated by weak topology of Banach space equal Borel $\sigma$-algebra for the norm topology?
This question is useful in showing that weak measurability in the sense of Bochner is equivalent to measurability in the usual way when we equip the Banach space with the Borel $\sigma$-algebra induced by the norm.
I can cover the case where the Banach space is separable.
Let $X$ our Banach space and denote the weak topology $\tau'$ and the $\sigma$-algebra it generates $\sigma(\tau')$. Since a closed ball is closed in the weak topology, all closed balls are in $\sigma(\tau')$. Let $B(x,r)$ an open ball. Then
$$B(x,r) = \bigcup_n \overline{B}(x,r-n^{-1}) \in \sigma(\tau').$$
Thus all open balls are in $\sigma(\tau')$, and since the open balls generate the Borel sigma algebra (using here separability), $\sigma(\tau')$ contains the Borel $\sigma$-algebra. The other inclusion is trivial, as the weak topology is weaker than the norm topology.
I'm unsure whether the result holds when $X$ is not separable.
Best Answer
Talagrand (1978) showed that the answer is "no" for the Banach space $l^\infty$.
Talagrand, Michel, Comparaison des boréliens d’un espace de Banach pour les topologies fortes et faibles, Indiana Univ. Math. J. 27, 1001-1004 (1978). ZBL0396.28007.
The question was asked by me in
Edgar, G. A., Measurability in a Banach space, Indiana Univ. Math. J. 26, 663-677 (1977). ZBL0361.46017.
and answered by Talagrand almost immediately.
I do know that the weak and norm Borel sets coincide if the space admits a Kadec norm. Corollary 2.4 in:
Edgar, G. A., Measurability in a Banach space. II, Indiana Univ. Math. J. 28, 559-579 (1979). ZBL0418.46034.