This question is followup to the previous similar question. There I was trying to find good sufficient condition for abstract preorder to be symmetric, but now, as I have found good formalization of my specific preorder, I'll better ask about it directly.

Let's say we have arbitrary finite measure space $\langle \Omega, \mathfrak{I}, \mathbb{P} \rangle$. For any functions $\rho : Y \rightarrow X$ and measurable $f : \Omega \rightarrow X$ with any finite $X$ and $Y$ we define $H_{\rho}(f) = \{\mathbb{P} \circ g^{-1} \mid f = \rho \circ g, g \text{ is measurable}\}$ – set of functions $Y \rightarrow \mathbb{R}_{\ge 0}$ characterizing internal structure of $f$ induced by $\rho$. Preorder $\precsim$ is defined on set of measurable functions $\{ f : \Omega \rightarrow X \}$ as follows – $f_1 \precsim f_2 \iff H_{\rho}(f_1) \subseteq H_{\rho}(f_2), \forall \rho$.

I believe that such preorder is symmetric ($f_1 \precsim f_2 \Leftrightarrow f_2 \precsim f_1, \forall f_1, f_2$) due to external reasons (it has respective game-theoretical interpretation), but I'm unable to prove it as defined. Can anyone help me, please?

UPDATE: Please, take note, $\forall \rho$ here means for all $\rho$ across every possible finite $Y$, not just one externally fixed.

UPDATE: To keep you guys interested and maybe bootstrap thoughts, I'll add some facts.

  1. $f_1 \precsim f_2 \Rightarrow \xi \circ f_1 \precsim \xi \circ f_2, \forall f_1, f_2, \xi$ – this follows trivially from definition of $H_{\rho}$.
  2. Consequentially, symmetry of $\precsim$ is easily provable for $f_2 = \pi \circ f_1, \pi : X \leftrightarrow X$. Because of $f_1 \precsim f_2 \Rightarrow \pi \circ f_1 \precsim \pi \circ f_2$ we can build infinite chain of $f_1 \precsim \pi \circ f_1 \precsim \pi \circ \pi \circ f_1 \precsim \ldots$. Combinatorics says that there must be finite cycle $\pi \circ ... \circ \pi$ synonymous to identity permutation, which leads to $f_2 = \pi \circ f_1 \precsim \pi \circ ... \circ \pi \circ f_1 = f_1$.
  3. Consequentially, symmetry of $\precsim$ is easily provable for measure spaces with finite algebra. In such case any measurable function $f_i$ can be derived from singular function $g$, that transforms each atom to its own element of codomain, as $f_i = \xi_i \circ g$. For such $g$ and its permutations $g \precsim \pi \circ g \Leftrightarrow \pi \circ g \precsim g \Leftrightarrow \mathbb{P} \circ g^{-1} = \mathbb{P} \circ (\pi \circ g)^{-1}$ which leads to $f_1 \precsim f_2 \Leftrightarrow f_2 \precsim f_1$ by definition of $H_{\rho}$.

Unfortunately, here my progress bumps into wall – going from this subcases to generic case seems to be beyond my knowlegde. And this is sad, because symmetry of preorder in question is a cornerstone of formal argumentation for very interesting new result in game theory. I would be very grateful even for slightest hints.

  • $\begingroup$ You should add measurability of $g$ in the definition of $H_\rho(f)$ (and perhaps delete or rethink the explanation after that definition). $\endgroup$ – Jochen Wengenroth May 8 at 7:16
  • $\begingroup$ If $f_1$ and $f_2$ have different distributions $\mathbb P^{f_j}=\mathbb P\circ f_j^{-1}$ the sets $H_\rho(f_j)$ are disjoint becaus any $Q\in H_\rho(f)$ satisfies $Q^\rho=\mathbb P^f$. Does this help? $\endgroup$ – Jochen Wengenroth May 8 at 7:19
  • $\begingroup$ @JochenWengenroth, added measurability of $g$, but I'm not sure what to do with explanation of $H_\rho$, as proper explanation would require bringing here pretty esoteric game-theoretical stuff, and I really want to avoid it. $\mathbb{P}^{f_1} = \mathbb{P}^{f_2}$ is a requirement for comparability of $f_1$ and $f_2$, as you well noted, but unfortunately it doesn't really help because roadblock here is a bit more complex - it's about quantified $\rho$ in both parts of $H_{\rho}(f_1) \subseteq H_{\rho}(f_2), \forall \rho \Leftrightarrow H_{\rho}(f_1) \supseteq H_{\rho}(f_2), \forall \rho$. $\endgroup$ – Doktor Diagoras May 8 at 7:51

I think I found a counterexample.

Let $Y=\{1\}$, $X=\Omega=\{1,2\}$, $f_1(\omega)=\omega$, $f_2(\omega)=1$.

Then for all $\rho:Y\to X$ there exists no $g$ such that $f=\rho \circ g$, thus $H_\rho(f_1)$ is empty and $f_1\precsim f_2$.

On the other hand, for $\rho = x\mapsto 1$ we have $H_\rho(f_2)=\{\mathbb P(\Omega)\}$. Thus $f_2\not\precsim f_1$.

  • $\begingroup$ I believe, we have a misunderstanding here. When I write $\forall \rho$, I mean for all $\rho$ across every possible finite $Y$, not just one externally fixed. I'll update my question. $\endgroup$ – Doktor Diagoras May 8 at 13:33

I did it, finally. Without going into much detail, it is possible to prove following fact - if there exists at least one pair of measurable $f_1, f_2 : \Omega \rightarrow X, f_1 \prec f_2 \equiv f_1 \precsim f_2 \cap \neg (f_1 \succsim f_2)$, then for any measurable $g_1 : \Omega \rightarrow Y$ it is possible to construct measurable $g_2 : \Omega \rightarrow Y$ for which $g_1 \prec g_2$. This is not possible though, because space of measurable functions with finite codomain is compact in Tychonoff topology, so for any partial order there must be at least one maximal element.

UPDATE: For people interested here is line of reasoning:

  1. I introduce pseudometric $d(f_1, f_2) = \mathbb{P}(f_1 \ne f_2)$ for space of $\Omega \rightarrow X$ and metric $d(\mu_1, \mu_2) = \max \lvert \mu_1 - \mu_2 \rvert$ for space of $Y \rightarrow \mathbb{R}_{\ge 0}$. Under this $H_\rho(f)$ is continuous in Hausdorff sense, so $H_\rho^{-1}(\mu) = \{\text{measurable } f \mid \mu \in H_\rho(f)\}$ is continuous too.
  2. Consequentially, if $f_1 \precsim f_2$, then for any measurable $g_1$ such that $f_1 = \rho \circ g_1$ it is possible to construct measurable $g_2$ such that $f_2 = \rho \circ g_2$ and $g_1 \precsim g_2$. Proof here uses triviality of building $g_2$ which $H_\rho(g_2)$ contains any finite subset of $H_\rho(g_1)$.
  3. Because of $f_1 \precsim f_2 \Rightarrow \rho \circ f_1 \precsim \rho \circ f_2$ previous statement holds for strict relation $\prec$ too.
  4. Supposing $f_1 \prec f_2 : \Omega \rightarrow X$, for any $g_1 : \Omega \rightarrow Y$ we build $h_1 = (f_1, g_1) : \Omega \rightarrow X \times Y$ and $\rho(x, y) = x, \xi(x, y) = y$. As $f_1 = \rho \circ h_1$ there exists measurable $h_2$ such that $f_2 = \rho \circ h_2$ and $h_1 \prec h_2$. Let's say $g_2 = \xi \circ h_2$ and because of $g_1 = \xi \circ h_1$ holds $g_1 \prec g_2$. Fortunately, this contradicts with compactness of space of measurable functions with finite codomain.
  • $\begingroup$ @LSpice, updated. Does it look like an answer now? $\endgroup$ – Doktor Diagoras May 24 at 0:18
  • $\begingroup$ I haven't read through the reasoning, but, yes, I think that providing the proof is appropriate. $\endgroup$ – LSpice May 24 at 1:32

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