# Again, proving that specific preorder on the set of measurable functions is symmetric

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.

• You should add measurability of $g$ in the definition of $H_\rho(f)$ (and perhaps delete or rethink the explanation after that definition). – Jochen Wengenroth May 8 at 7:16
• 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? – Jochen Wengenroth May 8 at 7:19
• @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 $H_{\rho}(f_1) \subseteq H_{\rho}(f_2), \forall \rho \Leftrightarrow H_{\rho}(f_1) \supseteq H_{\rho}(f_2), \forall \rho$. – Doktor Diagoras May 8 at 7:51

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 $$f_2\not\precsim f_1$$.
• 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. – Doktor Diagoras May 8 at 13:33