Let $\Omega=\Omega_1\times\Omega_2$ be a set equipped with a partial ordering that is inherited from partial orderings on $\Omega_1\times\Omega_2$. I.e. $(\omega_1,\omega_2)=\omega\leq\phi=(\phi_1,\phi_2)$ if and only if $\omega_1\leq\phi_1$ and $\omega_2\leq\phi_2$. An increasing set $I$ is a set where $\omega\in I$ and $\omega\leq\phi$ implies that $\phi\in I$.
Let $\nu=\nu_1\times\nu_2$ and $\mu=\mu_1\times\mu_2$ be product probability measures whose marginals are also probability measures. I am interested in when $\nu\preccurlyeq\mu$. This is a stochastic domination or stochastic ordering (more generally domination or ordering of measures). It is equivalent to $\nu(I)\leq\mu(I)$ for any increasing set $I$. Assume that the marginal sigma algebras have whatever properties they need to have, e.g. the sigma-algebra for $\nu_j$ is the same as for $\mu_j$ should be fine. Although, I would like to know the answer in a more general case.
Question: Is $\nu_1\times\nu_2\preccurlyeq\mu_1\times\mu_2$ equivalent to $\nu_1\preccurlyeq\mu_1$ and $\nu_2\preccurlyeq\mu_2$?
Here is my attempt at a proof.
$\nu_1\times\nu_2\preccurlyeq\mu_1\times\mu_2$ if and only if for an arbitrary increasing set $I\subset\Omega$ we have $(\nu_1\times\nu_2)(I)\preccurlyeq(\mu_1\times\mu_2)( I)$. Clearly $I_1\times\Omega_2$ is an increasing set when $I_1\subset\Omega_1$ is an increasing set. This gives us that $$\begin{aligned} \nu_1(I_1)&=\nu_1(I_1)\cdot\nu_2(\Omega_2)\\ &=(\nu_1\times\nu_2)(I_1\times\Omega_2)\\ &\leq(\mu_1\times\mu_2)(I_1\times\Omega_2)\\ &= \mu_1(I_1)\cdot\mu_2(\Omega_2)\\ &= \mu_1(I_1). \end{aligned} $$
So clearly $\nu_1\preccurlyeq\mu_1$ and we can likewise get that $\nu_2\preccurlyeq\mu_2$. Here I used that we are working with probability measures which makes me think the question is answered "no" for at least some product measures.
On the other hand, if we assume $\nu_j\preccurlyeq\mu_j$ for $j=1,2$. Then for an arbitrary increasing set $I\subset\Omega$ we can partition $I$ into "rectangular" sets $(A_1\times I_1)\cup (A_2\times I_2)\cup\cdots\cup (A_N\times I_N)$. We assume that each $A_j$ and $I_j$ are measurable by the appropriate $\nu_i, \mu_i$. The $A_j$ on the left partition the projection of $I$ onto $\Omega_1$ (denote that by $\pi_1(I)$), and the $I_j$ on the right are each increasing subsets of $\Omega_2$. Let's choose the partition fine enough so that
$$\nu(I)\approx \sum_{j=1}^N \nu(A_j\times I_j) ~ \text{ and } ~ \mu(I)\approx \sum_{j=1}^N \mu(A_j\times I_j)$$
Then we get
$$\begin{aligned} \nu(I) &=\int_{\pi_1(I)} \nu_2(I^x)d\nu_1(x)\\ &\approx\sum_{j=1}^N \nu(A_j\times I_j)\\ &=\sum_{j=1}^N \nu_1(A_j) \nu_2(I_j)\\ &\leq \sum_{j=1}^N \nu_1(A_j) \mu_2(I_j)\\ &\approx \int_{\pi_1(I)} \mu_2(I^x) d\nu_1(x)\\ &=\mu(I). \end{aligned}$$
Hence we conclude that $\nu_1\times\nu_2\preccurlyeq\nu_1\times\mu_2$. Similarly we can show that $\nu_1\times\mu_2\preccurlyeq\mu_1\times\mu_2$. This gives us $$\nu=\nu_1\times\nu_2\preccurlyeq\nu_1\times\mu_2\preccurlyeq\mu_1\times\mu_2=\mu.$$
I figure this argument could be more carefully formalized with $\epsilon$'s etc. My question is if the basic ideas seem correct. I know I hand-waved some things in there.
If this works, does it also work for infinite products? I'm interested in nice cases, say $\{0,1\}^{\mathbb N}$ with appropriate metric and/or topology, or more generally a Polish space or complete, separable metric space, something like that. My knowledge at this level is fairly rough...
The second part of your proof is not rigorous. Here is an alternative. First observe that for every measure $\nu$ on $(\Omega,\mathcal F)$, Fubini's theorem yields that $$\int_\Omega f \, d\nu = \int_\Omega \int_0^\infty {\bf 1}_{\{t \le f(x)\}} dt \, d\nu(x)= \int_0^\infty \int_\Omega {\bf 1}_{\{t \le f(x)\}} \, d\nu(x) \, dt \,.$$
Therefore, given measures $\nu,\mu$ on $(\Omega,\mathcal F)$, we have $\nu\preccurlyeq\mu$ iff all increasing positive measurable functions $f: \Omega \to \mathbb R$ satisfy $ \int_\Omega f \, d\nu \le \int_\Omega f \, d\mu$.
Now suppose that $\nu_i\preccurlyeq\mu_i$ on $\Omega_i$ for $i=1,2$. Given an increasing positive measurable function $g$ on $\Omega_1 \times \Omega_2$ endowed with the product $\sigma$-algebra, define for $x \in \Omega_1$ $$g_1(x):=\int_{\Omega_2} g_1(x,y) \, d\nu_2$$ and $$G_1(x):=\int_{\Omega_2} g_1(x,y) \, d\mu_2$$ so that $g_1 \le G_1$ are two positive increasing measurable functions on $\Omega_1$. We infer that $$ \int_{\Omega_1 \times \Omega_2} g(x,y) \, d\nu_2 \, d\nu_1 = \int_{\Omega_1} g_1(x) \, d\nu_1 \le \int_{\Omega_1} G_1(x) \, d\nu_1 $$ $$\le \int_{\Omega_1} G_1(x) \, d\mu_1 =\int_{\Omega_1 \times \Omega_2} g(x,y) \, d\mu_2\, d\mu_1 \,. $$