Does there exist a Borel measure or any valid measure on an infinite dimensional Banach space such that a bounded open set in this space has a positive measure ?
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1$\begingroup$ From memory only, so not 100% reliable, but: An exercise in P. Halmos, A Hilbert Space Problem Book, shows that it's impossible for a Hilbert space; so you should check that out first to see if the proof works for general Banach spaces. $\endgroup$– Zen HarperCommented Apr 14, 2011 at 7:03
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7$\begingroup$ As the question is written, if your space is separable, and so contains a countable dense subset, then can't you take a weighted sum of point masses? $\endgroup$– Matthew DawsCommented Apr 14, 2011 at 7:59
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1$\begingroup$ @Matthew, that might assign bounded open sets infinite measure. Of course, nothing in the problem rules that out, nor rules out the measure that assigns all non-empty sets measure $\infty$. $\endgroup$– LSpiceCommented Apr 14, 2011 at 17:41
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2$\begingroup$ @L Spice: Not if my measure is, say, $\sum_n 2^{-n} \delta_{x_n}$ where $(x_n)$ is a dense subset of $X$. Then the measure is itself finite... (Of course, as Anatoly mentions, perhaps the question meant to ask for an invariant measure...) $\endgroup$– Matthew DawsCommented Apr 14, 2011 at 18:55
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$\begingroup$ Oops, sorry; I missed ‘weighted’. $\endgroup$– LSpiceCommented Apr 16, 2011 at 14:11
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3 Answers
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The negative result mentioned in the comment by Zen Harper, is about invariant measures, non-existent on infinite-dimensional Banach spaces. If one does not require the invariance, there is no problem. See, for example, the calculation of the Gaussian measure of a ball in the paper
http://titan.math.udel.edu/~wli/papers/94-shifted-Kuelbs-Li-Linde.pdf
answered Apr 14, 2011 at 15:39
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It is a consequence of Riesz' Lemma that every open ball in an infinite dimensional normed space contains a disjoint sequence of smaller open balls. They all have the same measure under a translation invariant measure, so if the surrounding ball has finite measure, they all have measure zero. For separable spaces, this shows that every open set gets either measure 0 or $\infty$ under a translation invariant measure.
If you don't care about translation invariance, Wiener measure on the space of continuous function on [0,1] with starting value 0 should satisfy your condition.
answered Apr 14, 2011 at 22:05
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We can give a construction of a standard translation-invariant Borel measure in $R^N$(here $N$ denotes a set of all naturall numbers), which obtains the value one on the infinite-dimensional cube $[0;1[^N$. Actually, we are free from the demand of sigma-finiteness, because the space $R^N$ is covered by the uncountable family of pairwise disjoint shifts of $[0;1[^N$. Measures with above-mentioned properties are adopted as partial analogs of the Lebesgue measure in the infinite-dimensional topological vector space $R^N$. Partial analogs of the Lebesgue measure in general Banach spaces are assumed as translation-invariant Borel measures which obtain the numerical value one on the unit sphere or on the standard infinite-dimensional parallelepiped ( generated by any basis ).
The fundamental works of English mathematicians C. Rogers and D. Fremlin are devoted to problems of the existence of such measures in non-separable Banach spaces. I have considered the following problem posed by C. Rogers (1998):
Does there exist a such translation-invariant Borel measure in $\ell^{\infty}$ which obtains the numerical value one on the closed unite sphere?( here $\ell^{\infty}$ denotes a non-separable Banach space of all bounded real-valued sequences equipted with standard norm)
My result asserts that this question is not solvable within the the theory $ZF+DC$.
On the one hand, we can construct a "consistent" extension of the theory $ZF+DC$ where this question is solvable positivelly( such a theory is the so called "Solovay model")
On the other hand, we can construct a "consistent" extension of the theory $ZF+DC$ where this question is solvable negativelly ( such a theory is "ZF+AC+"there is no a measurable cardinal")
The proof of these facts can be found in
"G.R.Pantsulaia, On ordinary and standard products of infinite family of σ-finite measures and some of their applications Acta Mathematica Sinica, English Series (2011) 27: 477-496, March 01, 2011"
You have mentioned that in separable Banach spaces there is no a translation-invariant Borel measure which obtain a numerical value one on the unite ball. But if we consider a question asking whether there is a translation-invariant Borel measure in a separable Banach space which obtain a numerical value one on the infinite-dimensional parallelepiped ( generated by any Markushewicz basis, in particular, by Schauder basis ) then the answer to this question is yes.
