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This is perhaps a basic question, but I couldn't find a reference. Let $P = (X,\leq)$ be a poset. Given a probability measure $\mu$ on $P$ (with respect to the Borel $\sigma$-algebra generated by sets $S_p = \{q \in X: q \leq p\}$ for all $p \in X$), one can define its `CDF' $f_\mu(p) = \mu(S_p)$, the probability that a corresponding random element of $X$ is $\leq p$. This function is increasing with respect to the order $\leq$, and furthermore satisfies $$\lim_{n \to \infty} f_\mu(p_n) = 1 \text{ if $S_{p_n} \nearrow X$}$$ and $$\lim_{n \to \infty} f_\mu(p_n) = 0 \text{ if $S_{p_n} \searrow \emptyset$}.$$ The question: What conditions on either the function or the poset allow one to go backward and, given a function $f$ satisfying the above properties, realize it as the CDF (in the above sense) of a probability measure on our poset? I specifically want the case where $P$ is a lattice where each element covers a finite number of others, but may be countably infinite and have some infinite intervals.

In principle one can write down $\mu(\{p\})$ as an infinite linear combination of $f_\mu(q)$ for different $q$ by Möbius inversion, but given an $f$, it's not obvious to me when these linear combinations are nonnegative—and, for infinite posets, when they converge.

Edit: Thanks very much for the comments so far. Since from these it doesn't look like there's a nice well-known general theorem, let me say more about the specific setup I'm after and maybe this will be helpful. I take $X = \{(\lambda_1,\ldots,\lambda_d) \in (\mathbb{Z} \cup \{-\infty\})^d: \lambda_1 \geq \ldots \geq \lambda_d\}$ ('extended integer signatures'), and the ordering is given by the dominance order $$\lambda \geq \mu \Leftrightarrow \sum_{j=1}^i \lambda_j \geq \sum_{j=1}^i \mu_j \text{ for all }i=1,\ldots,d$$ (all sums with $-\infty$s in them are taken to be $-\infty$). This poset doesn't seem to satisfy the locally chain finite condition mentioned in the comments, but it's quite close to the poset of integer signatures $\{(\lambda_1,\ldots,\lambda_d) \in \mathbb{Z}^d: \lambda_1 \geq \ldots \geq \lambda_d\}$ with dominance order, which has all sorts of nice properties (e.g. all intervals are finite).

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  • $\begingroup$ What do you mean by "covers"? $\endgroup$ Commented May 3 at 17:50
  • $\begingroup$ In the real case the function should additionally be cadlag in the order topology. You might try this condition. $\endgroup$ Commented May 3 at 17:53
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    $\begingroup$ I don't think that there are any interesting non-linear orders for which these conditions will suffice. Consider e.g. the set $X$ being the set of all integral points in $[0,n]^2$ and $f(x,y)=\max(x,y)$ for $(x,y)\in X$. $\endgroup$ Commented May 3 at 17:53
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    $\begingroup$ I think you probably want some further finiteness conditions on the posets you're considering, Roger. For example, the set of rational numbers $\mathbb{Q}$ with its usual total order is a countable lattice where each element covers finitely many elements (in fact, no elements at all!) but is probably not what you had in mind. A possible additional condition you might want is "locally chain finite," i.e., no interval $[x,y]$ contains an infinite chain (although it may be infinite). $\endgroup$ Commented May 3 at 17:57
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    $\begingroup$ Thanks everyone, I've added more detail about the setting I'm after (I didn't put it in earlier because I figured there's probably a general theorem and didn't want to cloud the question with details, but it looks like maybe this isn't the case). @JulianHölz I think in my example one can get away with sequences rather than nets, though in more general settings I agree you're right. $\endgroup$ Commented May 3 at 18:53

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Not a complete answer but if you consider distributive lattices, I think this may be possible. If you have a distributive lattice with each element having a finite number of (co)atoms (i.e. elements immediately below it in the partial order), then the support of the Mobius function with one entry fixed is finite. One could then check for positivity (which I think is the big obstacle).

There is probably a lot more to say but I think the term to look for are valuations (I would recommend with Rota's papers with this term).

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