Comparing two sigma algebras in Hilbert spaces

Let $H$ be a non-separable Hilbert space. We denote $B$ by the sigma algebra generated by the norm topology in $H$. We also denote $B_{w}$ by the sigma algebra generated by the weak topology in $H$.

Question: Is $B$ the same as $B_w$?

Remark. When $H$ is separable, it is not difficult to see that they are the same.

The closed unit ball $M=\lbrace x\in H: \|x\|\le 1\rbrace$ does not belong to $B_w$. To see this you can assume that $H=\ell^2(I)=\lbrace (x_i)_{i\in I}: \sum_{i\in I}|x_i|^2=\sup \lbrace \sum_{i\in E} |x_i|^2: E\subset I \text { finite}\rbrace <\infty\rbrace$ for some uncountable index set $I$. Assuming $M\in B_w$ you would find a countable set $S\subseteq \ell^2(I)$ such that $M$ is contained in the $\sigma$-algebra $\mathscr S$ generated by $\lbrace \langle\cdot, x\rangle: x\in S\rbrace$ (because this is so for every element of $B_w$). Now, every $x\in S$ is supported in a countable set $J_x\subseteq I$. Then $J=\bigcup_{x\in S}J_x$ is countable and you get that for every $A\in \mathscr S$, $x\in A$ and $y\in \ell^2(I)$ with $x_j=y_j$ for all $j\in J$ you also have $y\in A$. However, $M$ clearly does not have this property even for $x=0$.
• There is also another point in your answer. Actually if we even assume that $B_w$ is the sigma algebra generated by the basic neighborhoods then (probably) your argument does not work. Because: you say that $M$ is contained in the sigma algebra $\mathscr S$. Note that $\langle.,x\rangle$ induces just the basic neighborhood centered at 0. – Ali Bagheri Jan 4 '16 at 13:43
• Okay, there is one point missing: The weak topology is the initial topology w.r.t all $\langle\cdot,x\rangle$. It is not clear that then $B_w$ is the initial $\sigma$-algebra w.r.t these functions. – Jochen Jan 4 '16 at 13:43