Let $(X, \leq)$ be a partial order and $\Sigma_X$ a $\sigma$-algebra on $X$. Is the set $\{(x, y) \in X\times X \mid x \leq y\}$ measurable with respect to the product $\sigma$-algebra?
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7$\begingroup$ You don't have any hypothesis connecting the order to the algebra, so this can't always be true. For example, nothing prevents $\Sigma_X$ from being the trivial algebra, which would be a counterexample if $X$ is nontrivial. Also, it isn't true in the reals $X=\mathbb{R}$, say, if we take the Borel sets and $\leq$ is a well-order. $\endgroup$– Joel David HamkinsCommented Sep 14, 2017 at 19:17
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$\begingroup$ In general topology, people typically assume separation axioms. I think it is therefore best to assume that every pair of points can be separated by a measurable set (a $\sigma$-algebra version of the separation axiom $T_{0}$) (it would be nice if this axiom were explicitly stated in the question to avoid trivialities). $\endgroup$– Joseph Van NameCommented Sep 14, 2017 at 19:28
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$\begingroup$ Of course - didn't think of the trivial one. $\endgroup$– daonCommented Sep 14, 2017 at 19:39
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2 Answers
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This can fail, even if you assume that the lower cones of the order are measurable with respect to the algebra $\Sigma$. For example, consider the real numbers $X=\mathbb{R}$ and let $\Sigma$ be the usual Borel algebra. Let's assume the continuum hypothesis, and let $\leq$ be a well-order of the reals in order type $\omega_1$. Thus, every lower cone $\{x\mid x\leq y\}$ is a countable set, which is therefore Borel. But the graph of the relation $\{(x,y)\mid x\leq y\}$ is not measurable, since all its horizontal sections are countable, but all its vertical sections are co-countable, which would violate Fubini's theorem.
answered Sep 14, 2017 at 19:34
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$\begingroup$ Interesting, thanks. Do you have any recommended reading on partial orders in topology/measure theory? $\endgroup$– daonCommented Sep 14, 2017 at 19:41
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$\begingroup$ It seems to be often considered, but I don't have any particular references to recommend. $\endgroup$ Commented Sep 14, 2017 at 19:52
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$\begingroup$ I would recommend the work of Guram Bezhanishvili and his colleagues. He has written quite a few papers on topological spaces with a compatible partial ordering. It turns out that much of the theory of general topology still holds when you have a compatible partial ordering on the topological spaces. Notions which generalize seamlessly include separation axioms, Stone duality, and the Stone-Cech compactification. $\endgroup$ Commented Sep 14, 2017 at 19:52
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Here is another counterexample.
I claim that the set $\{(x,y)\in X\times X|x\leq y\}$ is never measurable whenever $X$ has cardinality greater than the continuum. Suppose that $\{(x,y)\in X\times X|x\leq y\}$ is measurable. Then the set $\{(x,y)\in X\times X|x\geq y\}$ is also measurable. Therefore, the diagonal $\{(x,x)|x\in X\}=\{(x,y)\in X\times X|x\leq y\}\cap\{(x,y)\in X\times X|x\geq y\}$ is also measurable.
To see that the diagonal cannot be measurable, assume that $B\subseteq X\times X$. Then define $B_{x}=\{y\in X|(x,y)\in B\}$. Now define an equivalence relation $\simeq_{B}$ on $X$ where $x\simeq_{B}y$ if and only if $B_{x}=B_{y}$. Now let, $\mathcal{B}$ be the collection of all subsets of $X\times X$ where $\simeq_{B}$ has at most continuumly many equivalence classes. Then I claim that $\mathcal{B}$ is a $\sigma$-algebra. Clearly, $\mathcal{B}$ contains both $\emptyset$ and $X$. Furthermore, $\simeq_{B}=\simeq_{B^{c}}$, so $\mathcal{B}$ is closed under complementation. Now assume that $B_{n}\in\mathcal{B}$ for all $n\in\omega$ and let $B=\bigcup_{n}B_{n}$. Now let $\simeq$ be the equivalence relation on $X$ where $x\simeq y$ whenever $x\simeq_{B_{n}}y$ for all $n\in\omega$. Then define a mapping $i:X/\simeq\rightarrow \prod_{n\in\omega}X/\simeq_{B_{n}}$ by letting $i([x]_{\simeq})=([x]_{\simeq_{B_{n}}})_{n\in\omega}$. Then $i$ is an injective mapping, so since $|\prod_{n\in\omega}X/\simeq_{B_{n}}|\leq(2^{\aleph_{0}})^{\aleph_{0}}=2^{\aleph_{0}}$, we conclude that $X/\simeq$ has cardinality at most continuum. Now define a mapping $j:X/\simeq\rightarrow X/\simeq_{B}$ by $j([x]_{\simeq})=[x]_{\simeq_{B}}$. Then $j$ is surjective. Therefore $X/\simeq_{B}$ has cardinality at most continuum as well. Therefore, $B\in\mathcal{B}$, so $\mathcal{B}$ is a $\sigma$-algebra.
On the other hand, $\mathcal{B}$ contains all products $R\times S$ where $R,S\subseteq X$, but $\mathcal{B}$ does not contain the diagonal $\{(x,x)|x\in X\}$. Therefore, the product $\sigma$-algebra does not contain the diagonal either.
