# functional analysis – \$sigma\$-algebra generated by weak topology of Banach space equal Borel \$sigma\$-algebra

Yes. 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, $$sigma(tau’)$$ contains the Borel $$sigma$$-algebra. The other inclusion is trivial, as the weak topology is weaker than the norm topology.