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Consider the measurable space $2^{\mathbb R}$, equipped with the tensor-product $\sigma$-algebra. Famously, this space has a measurable structure which is not generated by a topology (see this answer).

Can you provide an example of a non-trivial probability measure on $2^{\mathbb R}$?

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Tom LaGatta
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3 Answers3

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Why can't you just take the product measure induced by the uniform measure on $\{0,1\}$ (or indeed by any other nontrivial measure on this two-element set)? I suppose my question really is: What exactly do you mean by non-trivial?

Andreas Blass
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    Product measures for an uncountable product are not, I think, an entirely standard topic; I wasn't even sure if they existed until after a little googling. "Trivial" probably means something like a sum of Dirac measures. – Qiaochu Yuan Nov 22 '13 at 00:36
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    @QiaochuYuan I confess to also not being as sure as I should have been. Being relatively old, I consulted my copy of Hewitt and Stromberg's "Real and Abstract Analysis" rather than Google, but with the same result: Uncountable products work just fine. (As a set theorist, I should have been more confident; the measure algebra associated to such a product is the standard way of forcing uncountably many random reals over a model of set theory.) – Andreas Blass Nov 22 '13 at 00:41
  • @AndreasBlass Actually, every normed measure algebra looks like a countable mixture of such coinflipping measures by Maharams classification theorem. – Michael Greinecker Nov 27 '13 at 19:21
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$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

Qiaochu Yuan
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  • Even if we assume the time line is homeomorphic to $\Bbb R$ somehow, flipping a coin $\aleph_0$ many times is an infinite task, but one can achieve that with immortality. Flipping $2^{\aleph_0}$ many coins will certainly amount to a condensation point in which an infinite amount of energy is released. I think we just figured out the big bang. :-) – Asaf Karagila Nov 22 '13 at 01:12
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    Yep; this follows from Kolmogorov's extension theorem. For a less trivial example, let the probability of the coin at $t$ be $1/(1+\exp(-B_t))$, where $B_t$ is a Brownian motion (run in both directions from $B_0=0$), say. – petrelharp Nov 22 '13 at 20:06
  • Thanks Qiaochu & @petrelharp. I think your two answers sufficiently resolve the question. Let $p : \mathbb R \to [0,1]$ be an arbitrary function (possibly non-measurable). Does $2^{\mathbb R}$ always support a product measure where the $t$th marginal is Bernoulli with probability $p(t)$? – Tom LaGatta Nov 22 '13 at 21:53
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    @Tom: yes, it seems there's no obstruction to just constructing the corresponding product measure. – Qiaochu Yuan Nov 22 '13 at 21:55
  • Qiaochu & @petrelharp, it dawns on me that I might have accepted this answer too soon. The motivation was to equip $2^{\mathbb R}$ with its tensor-product $\sigma$-algebra, which is not generated by a topology. Your answers use the Borel $\sigma$-algebra of the product topology, which seems to not actually answer the question. Can you speak to this point, particularly in light of Joseph Van Name's answer to this question? – Tom LaGatta Nov 26 '13 at 06:06
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    @Tom: if I understand correctly, and it's quite possible I haven't, the tensor product $\sigma$-algebra is strictly contained in the Borel $\sigma$-algebra of the product topology, so any measure I construct on the latter restricts to a measure on the former. (I admit I had initially completely forgotten about this distinction because I didn't know what you meant by "tensor product $\sigma$-algebra" and couldn't find a definition online.) – Qiaochu Yuan Nov 26 '13 at 06:22
  • @Qiaochu: thanks! I think you've resolved the issue. It's nice to see that there is an abundance of measures on this space, and not a dearth. For the record, I'll paste in the definition of tensor-product $\sigma$-algebra from Jochen Wengenroth's answer, along with the argument why it cannot be generated by a topology. http://mathoverflow.net/questions/87838/is-every-sigma-algebra-the-borel-algebra-of-a-topology/87888#87888 – Tom LaGatta Nov 27 '13 at 15:56

  • The underlying space is $\Omega= 2^{\mathbb R}$, that is the space of all indicator functions, and the $\sigma$-algebra is $\mathcal A = \bigotimes_{\mathbb R} \mathcal P$ where $\mathcal P$ is the power set of the two element set. It is generated by the system $\mathcal E$ of the "basic open sets" of the product topology (prescribed values in a finite number of points).

     – Tom LaGatta Nov 27 '13 at 15:56

  • This generator has the cardinality $c$ of the continuum and since the generated $\sigma$-algebra can be obtained in $\omega_1$ (transfinite) induction steps the cardinality of $\mathcal A$ is also $c$. On the other hand, if $\mathcal T$ is a topology with $\mathcal A=\sigma(\mathcal T)$ then $\mathcal T$ separates points (this should follow from the "good sets principle"), in particular, for two distinct points of $\Omega$ the closures of the corresponding singletons are distinct. Hence $\mathcal T$ has at least $|\Omega|=2^c$ elements.

     – Tom LaGatta Nov 27 '13 at 15:57
  • @TomLaGatta The tensor product $\sigma$-algebra is equal to the Baire $\sigma$-algebra. In general, on a compact Hausdorff space, Baire probability measures extend to Radon probability measures on the Borel sets (i.e. inner regular with respect to compact sets) and Radon probability measures on the Borel sets are equal iff they agree on Baire sets. – Robert Furber Oct 08 '18 at 16:04
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Instead of giving specific examples of probability measures on $2^{\mathbb{R}}$, I am going to give a couple characterizations of the tensor product $\sigma$-algebra on $2^{\mathbb{R}}$.

If $X$ is a topological space, then recall that set of the form $f^{-1}[\{0\}]$ for some continuous $f:X\rightarrow\mathbb{R}$ is called a zero set. The Baire $\sigma$-algebra on a topological space $X$ is the $\sigma$-algebra generated by the zero sets.

It turns out that if we give $2^{I}$ the product topology, then $2^{I}$ is a compact space and the Baire $\sigma$-algebra on $2^{I}$ is precisely the tensor product $\sigma$-algebra. In fact, if $X$ is a compact totally disconnected space, then the Baire $\sigma$-algebra on $X$ is generated by the clopen subsets of $X$. Therefore, by the Riesz representation theorem, the real-valued measures on the tensor product $\sigma$-algebra on $2^{I}$ are precisely the linear functionals on $C(2^{I})$. Of course, $C(2^{I})$ is the Banach space of continuous functions $2^{I}\rightarrow\mathbb{R}$. As a special case of this fact, the probability measures on the tensor product $\sigma$-algebra on $2^{I}$ are in a one-to-one correspondence with the positive linear functionals on $C(2^{I})$ (i.e. the functions $L:C(2^{I})\rightarrow\mathbb{R}$ with $L(f)\geq 0$ whenever $f\geq 0$).

Suppose that $X$ is a compact totally disconnected space. Let $B(X)$ denote the Boolean algebra of clopen subsets of $X$. Then the Baire probability measures on $X$ can be put into a one-to-one correspondence with finitiely additive probability measures on $B(X)$ as follows: If $\mu$ is a Baire probability measure on $X$, then $\mu|_{B(X)}$ is a finitely additive measure on $B(X)$. If $\nu$ is a finitely additive measure on $B(X)$, then we may extend $\nu$ to a Baire measure on $X$ by Caratheodory's Extension Theorem.

In particular, the probability measures on the tensor product $\sigma$-algebra on $2^{I}$ are in a one-to-one correspondence with the finitely additive probability measures on $B(2^{I})$. It should be noted that the Boolean algebras of the form $B(2^{I})$ are precisely the free Boolean algebras. Therefore, finding probability measures on the tensor product $\sigma$-algebra on $2^{I}$ simply amounts to finding finitely additive probability measures on free Boolean algebras.

Joseph Van Name
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