Let $\mu, \nu$ be probability measures on a partially ordered measurable space $(\Omega, \mathcal{F})$. We say that $\mu\preceq\nu$ ($\nu$ stochastic ally dominates $\mu$) if for every increasing event $A$ we have $$\mu(A)\le \nu(A).$$
With this definition, I now have a sequence of probability measures $\mu_n$ converging weakly to $\mu$ and another sequence of probability measures $\nu_n$ converging weakly to $\nu$. If I know that $\mu_n \preceq \nu_n$ for each $n$. Can I say that $\mu\preceq \nu$?
I have a feeling that it should be true. But I am not able to prove it. The problem is that under weak limit $\mu_n(A)$ need not converge to $\mu(A)$ for every increasing $A.$ I believe there is an alternate definition of stochastic domination in terms of integrals of increasing functions instead of increasing events. And obviously we can get that $$\int fd\mu \le \int fd\nu,$$ for every bounded continuous increasing function $f.$ Using some density argument, I think we should be able to push the inequality for every bounded increasing function.