3
$\begingroup$

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 $2^{\mathbb R}$?

$\endgroup$

3 Answers 3

8
$\begingroup$

$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

$\endgroup$
10
  • $\begingroup$ 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. :-) $\endgroup$
    – Asaf Karagila
    Commented Nov 22, 2013 at 1:12
  • 1
    $\begingroup$ 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. $\endgroup$
    – petrelharp
    Commented Nov 22, 2013 at 20:06
  • $\begingroup$ 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 $p(t)$? $\endgroup$ Commented Nov 22, 2013 at 21:53
  • 1
    $\begingroup$ @Tom: yes, it seems there's no obstruction to just constructing the corresponding product measure. $\endgroup$ Commented Nov 22, 2013 at 21:55
  • 1
    $\begingroup$ @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.) $\endgroup$ Commented Nov 26, 2013 at 6:22
10
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 22, 2013 at 0:36
  • 4
    $\begingroup$ @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.) $\endgroup$ Commented Nov 22, 2013 at 0:41
  • $\begingroup$ @AndreasBlass Actually, every normed measure algebra looks like a countable mixture of such coinflipping measures by Maharams classification theorem. $\endgroup$ Commented Nov 27, 2013 at 19:21
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks, Joseph. This is a really great response, and a welcome addition to the conversation. $\endgroup$ Commented Nov 22, 2013 at 21:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .