# Borel(X) = \sigma(X') for X non-separable

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.

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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.

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Thanks for the answer. I also downloaded your papers. That's very fascinating research. If you could give me the reference for $\ell^{\infty}$ that would be great. –  santker heboln May 13 '10 at 13:29
Gerald - Does the example of a Gaussian measure on $\ell^\infty$ prove the original question either way? My guess is that the cylindrical sigma algebra on $\ell^\infty$ is not the same thing as the Borel sigma algebra (otherwise, why even use the cylindrical sigma algebra in the example). But I'm not sure that the example implies this. –  George Lowther May 16 '10 at 16:12
I think it does since a Radon measure has a unique restriction to $\sigma(X')$. But I now have found a reference where it is shown directly that $\sigma(\ell^{\infty}') \not= \mathcal{B}(\ell^{\infty})$. See "Probability Distributions on Banach Spaces" by Vakhania et al, p. 23 - 24. –  santker heboln May 16 '10 at 20:16
Thanks Santker, that's a nice reference which answers your original question. I'm not sure what the significance of your comment about Radon measures is though (the example wasn't Radon). –  George Lowther May 17 '10 at 0:45