Products for probability theory using zero sets instead of open sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:27:12Z http://mathoverflow.net/feeds/question/87204 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87204/products-for-probability-theory-using-zero-sets-instead-of-open-sets Products for probability theory using zero sets instead of open sets Ricky Demer 2012-02-01T04:34:36Z 2012-02-01T04:34:36Z <p>(For all of this post, at least <a href="http://en.wikipedia.org/wiki/Axiom_of_countable_choice" rel="nofollow">Countable Choice</a> is assumed to hold.) <br><br><br> For all <a href="http://en.wikipedia.org/wiki/Tychonoff_space" rel="nofollow">Tychonoff spaces</a> $\langle X,\mathcal{T}\hspace{.06 in}\rangle$ :</p> <p>Define $\mathbf{Z}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ to be the set of subsets $S$ of $X$ such that there exists a <br> continuous $\: f : X\to (-\infty,\scriptsize+\normalsize\infty) \:$ such that <code>$\;\;\; \{x\in X : f(x)=0\} \; = \; S \;\;\;$</code>. <br> Define $\; \operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle) \;$ to be the sigma-algebra on $X$ generated by $\mathbf{Z}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$. <br> Define $\operatorname{UM}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ to be the set of subsets $S_0$ of $X$ such that <br> for all probability measures $\: \mu : \operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$, <br> there exists a member $S_1$ of $\operatorname{Ba}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ such that the symmetric difference between $S_0$ and $S_1$ is $\mu$-<a href="http://en.wikipedia.org/wiki/Null_set#Definition" rel="nofollow">null</a>. <br> It follows that $\operatorname{UM}(\langle X,\mathcal{T}\hspace{.06 in}\rangle)$ is a sigma-algebra on $X$.</p> <p><br></p> <p>Let $\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ be Tychonoff spaces. $\;\;$ Let $\: \mu_0 : \operatorname{UM}(\langle X,\mathcal{T}_0\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$ <br> and $\: \mu_1 : \operatorname{UM}(\langle Y,\mathcal{T}_1\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$ be probability measures. $\;\;$ Let $\mathcal{T}_2$ be the product topology <br> on $\: X\times Y \:$. $\;\;$ It follows that $\langle X\times Y,\mathcal{T}_2\rangle$ is a Tychonoff space and for all members $S_0$ of <br> $\operatorname{UM}(\langle X,\mathcal{T}_0\rangle)$, for all members $S_1$ of $\operatorname{UM}(\langle Y,\mathcal{T}_1\rangle)$, $\;\; S_0 \times S_1 \: \in \: \operatorname{UM}(\langle X\times Y,\mathcal{T}_2\rangle) \;\;$. <br> If there is a <a href="http://en.wikipedia.org/wiki/Measurable_cardinal#Measurable" rel="nofollow">measurable</a> <a href="http://en.wikipedia.org/wiki/Aleph_number" rel="nofollow">Aleph</a>, then there is an example where (*), defined as</p> <blockquote> <p>there exists a unique probability measure $\: \mu_2 : \operatorname{UM}(\langle X\times Y,\mathcal{T}_2\rangle) \to (-\infty,\scriptsize+\normalsize\infty) \:$ such that <br> for all members $S_0$ of $\operatorname{UM}(\langle X,\mathcal{T}_0\rangle)$, for all members $S_1$ of $\operatorname{UM}(\langle Y,\mathcal{T}_1\rangle)$, <br> $\mu_2(S_0 \times S_1) \; = \; \mu_0(S_0) \cdot \mu_1(S_1)$</p> </blockquote> <p>does not hold.</p> <p><br></p> <blockquote> <p>${}$1. $\:$ Does there provably-in-ZFC exist an example where (*) does not hold? <br> 1a. $\:$ If yes, how much of AC (beyond Countable Choice) does such a proof need to use? <br><br> ${}$2. <br> What amount of Choice (beyond Countable Choice) and/or <br> conditions on $\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ are known to force (*) to hold?</p> </blockquote> <p><br> I don't know any way full AC makes this easier, and the only condition I already know for 2 is:</p> <p>$\langle X,\mathcal{T}_0\rangle$ and $\langle Y,\mathcal{T}_1\rangle$ both have <a href="http://dantopology.wordpress.com/2009/11/14/spaces-with-countable-network/" rel="nofollow">countable networks</a> $\;\; \implies \;\;$ (*) <br><br></p>