8
$\begingroup$

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

$\endgroup$

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ May 13, 2010 at 13:29
  • $\begingroup$ 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. $\endgroup$ May 16, 2010 at 16:12
  • $\begingroup$ 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. $\endgroup$ May 16, 2010 at 20:16
  • $\begingroup$ 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). $\endgroup$ May 17, 2010 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.