An *ordered topological space* is a topological space $X$ equipped with a partial order $\leq$ which is closed as a subset of $X\times X$. By antisymmetry of $\leq$, it follows that the diagonal of $X$ is closed as well, so that $X$ is necessarily Hausdorff.


The [*stochastic order*](https://en.wikipedia.org/wiki/Stochastic_ordering) on the set of Radon probability measures on $X$ is given as follows. One defines $p\leq q$ if and only if $p(C) \leq q(C)$ for every closed upper set $C$. 



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> What is known about the antisymmetry of the stochastic order? Is it always antisymmetric? In other words, does $p(C) = q(C)$ for all closed upper sets $C$ imply $p = q$?
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Here's what I know: [Edwards (1978)](http://www.numdam.org/item/AIF_1978__28_4_53_0) has proven antisymmetry for compact $X$ (p.59). [Hiai, Lawson and Lim (2017)](https://arxiv.org/abs/1709.04187) have established it when $X$ is a certain type of cone in a Banach space (Theorem 4.3), using the $\pi$-$\lambda$ theorem. Which other results are there along these lines?



Antisymmetry of the stochastic order
I have been able to answer this question in the positive [for all ordered topological spaces](https://arxiv.org/abs/1810.06771).


The proof is completely elementary and unexpectedly simple.


