Characterization of pre-orders 
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*Let $X$ be an arbitrary set, and $\leq$ be a pre-order on $X$. Does there always exist $u:X\to \Bbb R$ such that $x'\leq x''$ iff $u(x') \leq u(x'')$? If this is not true in general, is that true under some additional assumptions?

*Let us endow now $X$ with some $\sigma$-algebra, and let $\mathcal P(X)$ be the set of all probability measures on $X$. We define a preorder on $\mathcal P(X)$ in terms of stochastic dominance: $p'\leq p''$ iff there exists a coupling $P\in \mathcal P(X\times X)$ with marginals $p',p''$ such that $P(\leq) = 1$.
Given that for $(X,\leq)$ there exists $u$ as in $(1)$, is that true that we can always refine $u$ to make sure that $p'\leq p''$ iff $\int u\;\mathrm dp'\leq \int u\;\mathrm dp''$?
 A: For (1), there is such a $u$ if and only if $\leq$ is complete and can be separated by a countable subset $Z \subset X$. In other words, for every $x$, $y \in X$ such that $x < y$, there exists $z \in Z$ such that $x \leq z \leq y$. For a counterexample in the case where the separation condition fails which doesn't involve set theory, the standard example in economics is the lexical ordering of the unit square: 
$(x_1, x_2) \prec (y_1, y_2) \iff (x_1 < y_1) \text{ or } (x_1 = y_1 \text{ and }  x_2 < y_2)$
If there were such a $u$, there would be uncountably many rationals.
For (2), I'm not sure what you mean by "refine $u$". In general, the stochastic dominance preorder will be incomplete, so there will be no such $u$ to start with. Representations of the form $p \leq p' \iff \int u \, dp \leq \int u \, dp'$ are so-called expected utility representations, and there are several ways a preorder on $P(X)$ can be complete and satisfy the stochastic dominance criterion without having an expected utility representation. 
This is all well-known, and will be covered in any standard textbook on utility theory.  
A: I think the question as stated is wrong - if we take any preorder with two incomparable elements, then clearly there is no such $u$. So maybe we want $u$ to preserve $<$, but not necessarily reflect it?

A counterexample to (1) which is maybe more interesting than an order which is just "too big": take the order $\omega_1$, the least uncountable ordinal.


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*Certainly $\omega_1$ is no bigger than $\mathbb{R}$ (at least, in ZFC), and may well be strictly smaller.

*On the other hand, there is no such map $u$: otherwise, for every $\alpha\in\omega_1$, we may find some rational number $q_\alpha$ between $u(\alpha)$ and $u(\alpha+1)$. Clearly $\alpha\not=\beta\implies q_\alpha\not=q_\beta$, but that implies that there are uncountably many rationals, contradiction.
By contrast, it's a fun exercise to show that every countable preorder does have such a $u$. Even more fun: every countable ordinal has some corresponding $u$ whose image in $\mathbb{R}$ is closed, whereas this is not in general true of countable preorders . . .
