Let $X$ be a Banach space, $X' = \mathcal{L}(X, \mathbb{K})$ its dual space. Denote by $\mathcal{B}(X)$ the $\sigma$-algebra of Borel sets and denote by $\sigma(X')$ the $\sigma$-algebra which is generated by all sets of the form $u^{-1}(C)$ for $u \in X'$ and $C \in \mathcal{B}(\mathbb{K})$.


For $X$ separable we have that


$\mathcal{B}(X) = \sigma(X')$ (\*)


see e.g. "Gaussian measures in Banach spaces" by Hui-Hsiung Kuo, p. 74 - 75.


Now the author of this book does not bother to discuss the case of $X$ non-separable.


In [1] is a halfway believable counterexample for $X = \ell^2(\mathbb{R})$.


I'm specifically interested in the case $X = \ell^{\infty}$. 
Does (\*) hold in this case and why or why not?


Thanks.


[1] <http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst;task=show_msg;msg=1533.0001.0001.0001>



Borel(X) = \sigma(X') for X non-separable 
If $I$ is uncountable, then in space $l^2(I)$ no countable set of functionals separates points. Consequently, for any set $A$ in the sigma-algebra generated by these functionals [the Baire sets for the weak topology, see reference below], if $0 \in A$, then an entire subspace is contained in $A$. So all elements of this sigma-algebra are unbounded. Thus this sigma-algebra is not all of the norm-Borel sets.


My papers on measurability in Banach space:  

Indiana Univ. Math. J. 26 (1977) 663--677  

Indiana Univ. Math. J. 28 (1979) 559--579


**edit**


For gaussian measures in Banach space, you *really* want the example
of Fremlin and Talagrand, "A Gaussian measure on $l^{\infty}$". Ann. Probab. 8 (1980), no. 6, 1192--1193. This gaussian measure on $l^\infty$ with the cylindrical sigma-algebra has total mass 1, yet every ball of radius 1 has measure 0.


