Normative Conspiracy Theory

Abstract

The author introduces a new model based on a correlated extension of normal-form games to describe the players’ behavior in environments that allow information asymmetry arising from different capabilities of private coordination of strategies.

APPENDIX

To prove the theorem on isomorphic spaces, we have to introduce additional tools.

Definition A.1.

A refinement of a set $X = X^1 \times \ldots \times X^m$ of outcomes to a finite set $Y = Y^1 \times \ldots \times Y^m$ of outcomes is any mapping $\rho = (\rho ^1, \ldots , \rho ^m)$, where each component $\rho ^a$ takes $Y^a$ to $X^a$.

Using refinements, one can specify connections between partitions with different codomains. If partitions $f : \Omega \rightarrow X$ and $g : \Omega \rightarrow Y$ are such that $f = \rho \circ g$, then $f^{-1}(x) = \bigcup _{y \in \rho ^{-1}(x)} g^{-1}(y), \forall x \in X$. In this case, $f$ is said to be refinable to $g$.

Partitions of one and the same space can be combined. For example, of the partitions $g_i : \Omega \rightarrow Y_i = Y_i^1 \times \ldots \times Y_i^m, i={1,\ldots ,n}$, one can construct their combination $g_1 \diamond \ldots \diamond g_n : \Omega \rightarrow Y_{(n)}$, where $Y_{(n)}^a = Y_1^a \times \ldots \times Y_n^a$ and $(g_1 \diamond \ldots \diamond g_n)^a(\omega ) = (g_1^a(\omega ), \ldots , g_n^a(\omega )), \forall \omega \in \Omega$, $a ={1,\ldots ,m}$. This combination of partitions is related to its components via refinement-projections,

$$ g_i = \pi _i \circ (g_1 \diamond \ldots \diamond g_n),\quad \pi _i^a(x_1^a, \ldots , x_n^a) = x_i^a. $$

Refinements with a common codomain can be combined in a similar way. For example, from the refinements $\rho _i : Y_i \rightarrow X, i={1,\ldots ,n}$, one can construct a combination $\rho _1 \wr \ldots \wr \rho _n : Y_{[n]} \rightarrow X$, where $Y_{[n]}^a = \{(y_1^a, \ldots , y_n^a) \in Y_{(n)}^a \mid \rho _1^a(y_1^a) = \ldots = \rho _n^a(y_n^a)\}$, $a ={1,\ldots ,m}$, and $\rho _1 \wr \ldots \wr \rho _n$ coincides on the domain of itself with all $\rho _i \circ \pi _i$. Note that

$$ f = \rho _i \circ g_i,\quad i={1,\ldots ,n} \;\Leftrightarrow \; f = (\rho _1 \wr \ldots \wr \rho _n) \circ (g_1 \diamond \ldots \diamond g_n).$$

Definition A.2.

In a correlation space $\Phi = \langle A, \Omega , \mathfrak {I}^a, \mathbb {P}, a \in A \rangle$, the structure of partition $f : \Omega \rightarrow X$ generated by a refinement $\rho : Y \rightarrow X$ is the set $H_{\Phi ,\rho }(f) = \{\mathbb {P} \circ g^{-1} \mid g \models \Phi , f = \rho \circ g\}$ consisting of the measures $\mu : Y \rightarrow \mathbb {R}_{\ge 0}$. We also set $H_{\Phi ,\rho }^{-1}(\mu ) = \{f \models \Phi \mid \mu \in H_{\Phi ,\rho }(f)\}$.

Lemma A.1.

For all $\rho : Y \rightarrow X$ and $\mu : Y \rightarrow \mathbb {R}_{\ge 0}$ , the set $H_{\Phi ,\rho }^{-1}(\mu ) \subseteq X^{\Omega }$ is compact in the semimetric

$$ \operatorname {dis}(f_1, f_2) = \frac {1}{2} \sum _{x \in X} \Big | \mathbb {P}\big (f_1^{-1}(x)\big ) - \mathbb {P}\big (f_2^{-1}(x)\big ) \Big |.$$

Proof. Let us restate $H_{\Phi ,\rho }^{-1}(\mu ) = \rho \circ H_{\Phi }^{-1}(\mu )$ by defining $H_{\Phi }^{-1}(\mu ) = \{g \models \Phi \mid \mathbb {P} \circ g^{-1} = \mu \}$. First, we prove the compactness of $H_{\Phi }^{-1}(\mu )$ by introducing $\operatorname {dis}(g_1, g_2)$ by analogy with $\operatorname {dis}(f_1, f_2)$. The semimetric $\operatorname {dis}$ is completely bounded, because $\operatorname {dis}(g_1, g_2) = d(\mu ^Y_1, \mu ^Y_2)$, where $\mu ^Y_k = \mathbb {P} \circ g_k^{-1}, k=1,2$, and the space of probability measures is completely bounded on any finite set. The closedness of $H_{\Phi }^{-1}(\mu )$ follows in an obvious way from the equivalence $\operatorname {dis}(g_1, g_2) = 0 \Leftrightarrow \mathbb {P} \circ g_1^{-1} = \mathbb {P} \circ g_2^{-1}$. Thus, $H_{\Phi }^{-1}(\mu )$ is compact in the semimetric $\operatorname {dis}$. Let us prove the continuity of the mapping $\rho \ \circ : Y^\Omega \rightarrow X^\Omega$ by differently expressing the same semimetric,

$$ \operatorname {dis}(f_1, f_2) = 1 - \sum _{x \in X} \min \Big [\mathbb {P}\big (f_1^{-1}(x)\big ), \mathbb {P}\big (f_2^{-1}(x)\big )\Big ].$$

Now let $f_1 = \rho \circ g_1$ and $f_2 = \rho \circ g_2$,

$$ \begin {aligned} \operatorname {dis}(\rho \circ g_1, \rho \circ g_2) &= 1 - \sum _{x \in X} \min \Big [\mathbb {P}\big (g_1^{-1}\big (\rho ^{-1}(x)\big )\big ), \mathbb {P}\big (g_2^{-1}\big (\rho ^{-1}(x)\big )\big )\Big ] \\ &= 1 - \sum _{x \in X} \min \left [\,\sum _{y \in \rho ^{-1}(x)} \mathbb {P}\big (g_1^{-1}(y)\big ), \sum _{y \in \rho ^{-1}(x)} \mathbb {P}\big (g_2^{-1}(y)\big )\right ] \\ &\le 1 - \sum _{x \in X} \sum _{y \in \rho ^{-1}(x)} \min \Big [\mathbb {P}\big (g_1^{-1}(y)\big ), \mathbb {P}\big (g_2^{-1}(y)\big )\Big ] \\ &= 1 - \sum _{y \in Y} \min \Big [\mathbb {P}\big (g_1^{-1}(y)\big ), \mathbb {P}\big (g_2^{-1}(y)\big )\Big ] = \operatorname {dis}(g_1, g_2). \end {aligned}$$

The mapping $\rho \ \circ$ is continuous, because $\operatorname {dis}(\rho \circ g_1, \rho \circ g_2) \le \operatorname {dis}(g_1, g_2)$. Since continuous mappings preserve compactness [4, p. 199], it follows that $H_{\Phi ,\rho }^{-1}(\mu ) = \rho \circ H_{\Phi }^{-1}(\mu )$ is compact in the semimetric $\operatorname {dis}$. $\quad \blacksquare$

Definition A.3.

A partition $f_2 \models \Phi _2$ is called an exact image of a partition $f_1 \models \Phi _1$ (hereinafter $f_1 \precsim f_2$) if their codomains coincide ($X_1 = X_2 = X$) and $H_{\Phi _1,\rho }(f_1) \subseteq H_{\Phi _2,\rho }(f_2)$ for all refinements $\rho$ with the same codomain. The set of all exact images will be denoted in the sequel by $\widehat {\Phi }_2(f_1) = \{f_2 \models \Phi _2 \mid f_1 \precsim f_2\}$.

The relation $f_1\precsim f_2$ can be understood as follows: no matter into what measurable parts we divide the components of the partition $f_1$, the corresponding components in the partition $f_2$ can always be divided into parts equal to them in measure.

Remark A.1.

Obviously, $f_1 \precsim f_2 \wedge f_2 \precsim f_3 \Rightarrow f_1 \precsim f_3$.

Lemma A.2.

Assume that partitions $g_1 : \Omega _1 \rightarrow Y$ and $g_2 : \Omega _2 \rightarrow Y$ in correlation spaces $\Phi _1$ and $\Phi _2$ are such that $g_1 \precsim g_2$ . Then $\rho \circ g_1 \precsim \rho \circ g_2$ for all refinements $\rho : Y \rightarrow X$ .

Proof. Take any $\rho _* : Y_* \rightarrow X$ and $\mu \in H_{\Phi _1,\rho _*}(\rho \circ g_1)$. By the definition of the structure of partition, $\exists g_{1*}: \rho _* \circ g_{1*} = \rho \circ g_1, \mathbb {P}_1 \circ g_{1*}^{-1} = \mu$, and we need to prove, by the definition of the exact image, that $\exists g_{2*}: \rho _* \circ g_{2*} = \rho \circ g_2, \mathbb {P}_2 \circ g_{2*}^{-1} = \mu$. Consider a combination $g_{1+} = g_1 \diamond g_{1*}$, where $g_1 = \pi \circ g_{1+}$ and $g_{1*} = \pi _* \circ g_{1+}$. Here $g_{1+} : \Omega _1 \rightarrow Y_+, Y_+^a = Y^a \times Y_*^a, a={1,\ldots ,m}$. By the definition of the structure of partition, $\mathbb {P}_1 \circ g_{1+}^{-1} \in H_{\Phi _1,\pi }(g_1)$, and hence, since $g_1 \precsim g_2$, there exists a $g_{2+} : \Omega _2 \rightarrow Y_+$ such that $\mathbb {P}_1 \circ g_{1+}^{-1} = \mathbb {P}_2 \circ g_{2+}^{-1} \in H_{\Phi _2,\pi }(g_2)$; i.e., $\pi \circ g_{2+} = g_2$. This obviously implies that $\mathbb {P}_2 \circ (\pi _* \circ g_{2+})^{-1} = \mathbb {P}_1 \circ (\pi _* \circ g_{1+})^{-1}$ as well, and hence $g_{2*} = \pi _* \circ g_{2+}$ is the desired partition. $\quad \blacksquare$

Lemma A.3.

Assume that partitions $f_1 : \Omega _1 \rightarrow X$ and $f_2 : \Omega _2 \rightarrow X$ in correlation spaces $\Phi _1$ and $\Phi _2$ are such that $f_1 \precsim f_2$ . Then for each refinement $\rho : Y \rightarrow X$ and each partition $g_1 : \Omega _1 \rightarrow Y$ such that $f_1 = \rho \circ g_1$ there exists a partition $g_2 : \Omega _2 \rightarrow Y$ such that $f_2 = \rho \circ g_2$ and $g_1 \precsim g_2$ .

Proof. Let us state the desired assertion as $\exists g_2 \in \widehat {\Phi }_2(g_1) : f_2 = \rho \circ g_2$ and express $\widehat {\Phi }_2$ via the structure of partitions as

$$ \widehat {\Phi }_2(g_1) = \bigcap _{\forall Z, \xi : Z \rightarrow Y, \mu \in H_{\Phi _1,\xi }(g_1)} H_{\Phi _2,\xi }^{-1}(\mu ).$$

By Lemma A.1 the set $\widehat {\Phi }_2(g_1)$ is the intersection of a family of compact sets. Consequently, to prove that it contains the element $g_2 : f_2 = \rho \circ g_2$, it suffices to prove that such an element is contained in the intersection of each finite subfamily of the same compact sets,

$$ \exists g_{2*} \in \bigcap _{i=1}^n H_{\Phi _2,\xi _i}^{-1}(\mu _i) : f_2 = \rho \circ g_{2*}, $$

where $\xi _i : Z_i \rightarrow Y$ are arbitrary refinements with arbitrary domains $Z_i$ and the $\mu _i \in H_{\Phi _1,\xi _i}(g_1)$ are chosen arbitrarily as well.

By the definition of the structure of partition, $\exists h_{1,i} \models \Phi _1 : g_1 = \xi _i \circ h_{1,i}, \mathbb {P} \circ h_{1,i}^{-1} = \mu _i$. Let us construct their combination $h_1 = h_{1,1} \diamond \dots \diamond h_{1,n}$, where $h_{1,i} = \pi _i \circ h_1$, and denote $\xi = \xi _1 \wr \dots \wr \xi _n$. By the definition of the exact mapping, $\exists h_2 \models \Phi _2 : f_2 = \rho \circ \xi \circ h_2, \mathbb {P}_1 \circ h_1^{-1} = \mathbb {P}_2 \circ h_2^{-1}$, and hence we can take $g_{2*} = \xi \circ h_2$. By construction, $f_2 = \rho \circ g_{2*}$ and $\mathbb {P}_1 \circ h_{1,i}^{-1} = \mathbb {P}_1 \circ (\pi _i \circ h_1)^{-1} = \mathbb {P}_2 \circ (\pi _i \circ h_2)^{-1} = \mathbb {P}_2 \circ h_{2,i}^{-1}$; consequently, $g_{2*}$ is the desired partition. $\quad \blacksquare$

Corollary A.1.

If correlation spaces satisfy $\Phi _1 \precsim \Phi _2$ , then for each partition $f_1 \models \Phi _1$ there exists an $f_2 \models \Phi _2$ such that $f_1 \precsim f_2$ .

Corollary A.2.

Lemmas A.2 and A.3 and Corollary A.1 remain valid for the strict relation $f_1 \prec f_2 \equiv f_1 \precsim f_2 \cap \neg (f_1 \succsim f_2)$ .

Lemma A.4.

One has $f_1 \precsim f_2 \Leftrightarrow f_1 \succsim f_2$ for any partitions of one and the same correlation space.

Proof. Assume the contrary: there exists an $f_1 \prec f_2$ with the codomain $X$. The trivial refinement $\theta (x) = (0, \ldots , 0), \forall x \in X$ obviously yields $\theta \circ f_1 = \theta \circ f_2$. This contradicts $\theta \circ f_1 \prec \theta \circ f_2$, which follows from Lemma A.2. $\quad \blacksquare$

Corollary A.3.

One has $f_1 \precsim f_2 \Leftrightarrow f_1 \succsim f_2$ for any partitions of isomorphic correlation spaces.

Proof of the theorem on isomorphic spaces (Theorem 3.1). Take an arbitrary profile $\mathbf {s}_1$ of strategies in the game $\Gamma | \Phi _1$. Obviously, this profile is a partition of the correlation space $\Phi _1$. By Corollary A.1, there exists a partition $\mathbf {s}_2$ of the correlation space $\Phi _2$ such that $\mathbf {s}_1 \precsim \mathbf {s}_2$, and in a similar way, $\mathbf {s}_2$ is also a profile of strategies in the game $\Gamma | \Phi _2$. Let us prove the embeddings in both directions: (1) $U_{\Gamma | \Phi _1}^{A_*}(\mathbf {s}_1) \subseteq U_{\Gamma | \Phi _2}^{A_*}(\mathbf {s}_2)$ and (2) $U_{\Gamma | \Phi _1}^{A_*}(\mathbf {s}_1) \supseteq U_{\Gamma | \Phi _2}^{A_*}(\mathbf {s}_2)$ for each cabal $A_*$ of players:

• 1. Consider an arbitrary profile $\mathbf {s}_{1*} \models \Phi _1$ different from $\mathbf {s}_1$ by the strategies of the cabal $A_*$. Denote $\mathbf {s}_{1+} = \mathbf {s}_1 \diamond \mathbf {s}_{1*}$, where $\mathbf {s}_1 = \pi \circ \mathbf {s}_{1+}$ and $\mathbf {s}_{1*} = \pi _* \circ \mathbf {s}_{1+}$. By the definition of exact image, we have $H_{\Phi _1,\pi }(\mathbf {s}_1) \subseteq H_{\Phi _2,\pi }(\mathbf {s}_2)$; i.e., $\exists \mathbf {s}_{2+} \models \Phi _2 : \mathbb {P}_1 \circ \mathbf {s}_{1+} = \mathbb {P}_2 \circ \mathbf {s}_{2+}, \mathbf {s}_2 = \pi \circ \mathbf {s}_{2+}$. By construction, $\mathbf {s}_{2*} = \pi _* \circ \mathbf {s}_{2+}$ is different from $\mathbf {s}_2$ by the moves of the same players that distinguish $\mathbf {s}_{1*}$ from $\mathbf {s}_1$, and $\mathbb {P}_1 \circ \mathbf {s}_{1*}^{-1} = \mathbb {P}_2 \circ \mathbf {s}_{2*}^{-1}$, and hence, in a similar way, $u^a(\mathbf {s}_{1*}) = u^a(\mathbf {s}_{2*})$. By virtue of arbitrariness of the choice of $\mathbf {s}_{1*}$, this implies that $U_{\Gamma | \Phi _1}^{A_*}(\mathbf {s}_1) \subseteq U_{\Gamma | \Phi _2}^{A_*}(\mathbf {s}_2)$.

• 2. Since $\mathbf {s}_1 \succsim \mathbf {s}_2$ by Corollary A.3, the reasoning in the previous item is applicable in both forward and backward directions. $\quad \blacksquare$

To prove the theorem on the isomorphism of conspiracy spaces, we need several lemmas.

Lemma A.5.

For each countable family $\mathfrak {F}$ of sets, there exists a chain $\mathfrak {T}$ of sets such that $\sigma (\mathfrak {F}) = \sigma (\mathfrak {T})$ .

Proof. Let $\mathfrak {F} = \{F_1, F_2, \ldots \}$. Let us construct by induction a sequence of chains $(\mathfrak {T}_i)$ where each next chain incorporates the previous one and $\sigma (\mathfrak {T}_i) = \sigma (\{F_1, \ldots , F_i\})$. For the base case we take $\mathfrak {T}_1 = \{F_1\}$. The induction step is as follows: let $\mathfrak {T}_{i-1} = \{T_1, \ldots , T_n\}, T_1 \subset \ldots \subset T_n$, and $\sigma (\mathfrak {T}_{i-1}) = \sigma (\{F_1, \ldots , F_{i-1}\})$. We decompose the next element $\mathfrak {F}$ into disjoint disjunctions, $F_i = (F_i \cap T_1) \cup (F_i \cap T_2 \setminus T_1) \cup \ldots \cup (F_i \cap T_n \setminus T_{n-1}) \cup (F_i \setminus T_n)$. In this notation, the $j$th disjunction is embedded in the corresponding difference $T_j \setminus T_{j-1}$ of neighboring chain elements. Consequently, to generate it, it suffices to augment $\mathfrak {T}_{i-1}$ with the set $T_{j-} = F_i \cap T_j \cup T_{j-1}$, which preserves the chain structure, because $T_{j-1} \subseteq T_{j-} \subseteq T_j$. Thus, to produce the entire $F_i$, we set

$$ \mathfrak {T}_i = \mathfrak {T}_{i-1} \cup \left \{ \begin {aligned} &F_i \cap T_1,\\ &F_i \cap T_2 \cup T_1,\\ &\cdots \\ &F_i \cap T_n \cup T_{n-1},\\ &F_i \cup T_n \end {aligned} \right \}.$$

Let us show that the limit of the sequence $(\mathfrak {T}_i)$ is the desired chain. Indeed, each element of $\sigma (\mathfrak {F})$ is a countable union of finite intersections of the sets $F_i$. Therefore,

$$ \sigma (\mathfrak {F}) = \bigcup _{i=1}^{\infty } \sigma (\{F_1, \ldots , F_i\}) = \bigcup _{i=1}^{\infty } \sigma (\mathfrak {T}_i) = \sigma (\mathfrak {T}).\quad \blacksquare$$

Lemma A.6.

The maximum chain of measurable sets in an atomless space generates an atomless $\sigma$ -algebra.

Proof. Let the maximum chain $\mathfrak {T}$ of measurable sets of the atomless space $\langle \Omega , \mathfrak {B}, \mathbb {P} \rangle$ generate an algebra $\sigma (\mathfrak {T})$. Let us prove that for each $B \in \sigma (\mathfrak {T})$ of measure $\mathbb {P}(B) > 0$ there exists a $B^{\prime } \in \sigma (\mathfrak {T})$ such that $B^{\prime } \subset B$ and $\mathbb {P}(B) > \mathbb {P}(B^{\prime }) > 0$. To this end, obviously, it suffices to prove that in the chain $\mathfrak {T}$ there exists a set $T$ such that $0 < \mathbb {P}(T \cap B) < \mathbb {P}(B)$. Consider the sets

$$ \underline {T} = \bigcup _{T_- \in \mathfrak {T} : \mathbb {P}(T_- \cap B) = 0} T_- \quad \text {and}\quad \overline {T} = \bigcap _{T_+ \in \mathfrak {T} : \mathbb {P}(T_+ \cap B) = \mathbb {P}(B)} T_+,$$

which are nested $\underline {T} \subset \overline {T}$ by construction, so that $\mathbb {P}(\overline {T}) - \mathbb {P}(\underline {T}) \ge \mathbb {P}(B)$. Since the chain $\mathfrak {T}$ is maximal in the atomless space, it follows that there exists a $T_0 \in \mathfrak {T}$ such that $\underline {T} \subset T_0 \subset \overline {T}$. Since $\underline {T} \subset T_0 \Rightarrow \mathbb {P}(T_0 \cap B) > 0$ and $T_0 \subset \overline {T} \Rightarrow \mathbb {P}(T_0 \cap B) < \mathbb {P}(B)$, we conclude that $T_0$ is the desired set. $\quad \blacksquare$

Definition A.4.

For any families of measurable sets $\mathfrak {T} \subseteq 2^\Omega$ and measures $\mathbb {P} : \mathfrak {T} \rightarrow \mathbb {R}_{\ge 0}$, we define a mapping $\operatorname {mim}\langle \mathfrak {T}, \mathbb {P}\rangle : \Omega \rightarrow \mathbb {R}_{\ge 0}$ referred to as the least measure of inclusion and calculated by the formula $\operatorname {mim}\langle \mathfrak {T}, \mathbb {P}\rangle (\omega ) = \inf \{\mathbb {P}(T) \mid \omega \in T \in \mathfrak {T}\}$.

Lemma A.7.

If $\mathfrak {T} \subset 2^\Omega$ is a chain of sets that generates an atomless $\sigma$ -algebra, then $\mathbb {P} \circ \operatorname {mim}\langle \mathfrak {T}, \mathbb {P}\rangle ^{-1}$ coincides with the Lebesgue measure on the interval $[0,\mathbb {P}(\Omega )]$ .

Proof. Since $\mathfrak {T}$ is a chain, we have $\omega \in T \Leftrightarrow \operatorname {mim}\langle \mathfrak {T}, \mathbb {P}\rangle (\omega ) \le \mathbb {P}(T), \forall \omega \in \Omega , T \in \mathfrak {T}$. Since, in addition, $\mathfrak {T}$ generates an atomless $\sigma$-algebra, we conclude that for each $0 < t < \mathbb {P}(\Omega )$ there exists a $T \in \mathfrak {T}$ such that $\mathbb {P}(T) = t$. Consequently, the function $\operatorname {mim}\langle \mathfrak {T}, \mathbb {P}\rangle$ maps the sets $T \in \mathfrak {T}$ onto the intervals $[0, \mathbb {P}(T)]$, which obviously implies the desired assertion. $\quad \blacksquare$

Lemma A.8.

Let $\langle \Omega , \mathfrak {B}, \mathbb {P} \rangle$ be an atomless probability space with a $\sigma$ -algebra decomposable into $n$ atomless components $\mathfrak {B} = \sigma (\mathfrak {B}_1 \cup \ldots \cup \mathfrak {B}_n)$ such that all events from different components are jointly independent; i.e., $\mathbb {P}(B_1 \cap \ldots \cap B_n) = \mathbb {P}(B_1) \ldots \mathbb {P}(B_n)$ for any $B_i \in \mathfrak {B}_i, i={1,\ldots ,n}$ . Then any measurable function $f : \Omega \rightarrow X$ with a finite codomain is representable in the form $f = \varphi \circ \mathfrak {r}$ , where $\mathfrak {r} : \Omega \rightarrow [0,1]^n$ is such that $\mathbb {P} \circ \mathfrak {r}^{-1}$ coincides with the Lebesgue measure and $\varphi : [0,1]^n \rightarrow X$ is a Borel function.

Proof. Consider the inverse function $f^{-1} : X \rightarrow \mathfrak {B}$. By virtue of the decomposability of $\mathfrak {B}$, it can be represented as the limit of a sequence of conjunctions,

$$ f^{-1}(x) = \bigcup _{j=1}^\infty F_1^j(x) \cap \ldots \cap F_n^j(x), \; F_i^j : X \rightarrow \mathfrak {B}_i.$$

Consider the families $\mathfrak {F}_i = \{F_i^j(x) \mid j \in \mathbb {N}, x \in X\}$ of sets and note that $f$ is measurable according to $\sigma (\mathfrak {F}_1 \cup \ldots \cup \mathfrak {F}_n)$. By Lemma A.5, there exist chains of sets $\mathfrak {T}_i \subset \mathfrak {B}_i$ such that $\sigma (\mathfrak {F}_i) = \sigma (\mathfrak {T}_i)$. According to the Hausdorff maximum principle, each such chain is embedded in a maximal chain $\overline {\mathfrak {T}}_i \subset \mathfrak {B}_i$ generating an atomless $\sigma$-algebra by Lemma A.6. Let us construct the desired $\mathfrak {r} = (\operatorname {mim}\langle \overline {\mathfrak {T}}_1, \mathbb {P}\rangle , \ldots , \operatorname {mim}\langle \overline {\mathfrak {T}}_n, \mathbb {P}\rangle )$ and $\varphi = f \circ \mathfrak {r}^{-1}$. The necessary properties are observed by construction. $\quad \blacksquare$

Proof of Theorem 4.1. Let us apply the previous lemma to an arbitrary conspiracy space $\Phi _1$ of the structure $\mathfrak {A} = \{A_1, \ldots , A_n\}$ using the secrets of the cabals of conspirators for the respective components of the decomposition $\mathfrak {B}_1, \ldots , \mathfrak {B}_n$ of the $\sigma$-algebra. This yields a decomposition $f_1 = \varphi \circ \mathfrak {r}$ for each partition $f_1 \models \Phi _1$. In any other conspiracy space $\Phi _2$ of the same structure $\mathfrak {A}$, the relevant partition $f_2 \models \Phi _2$ is constructed in a similar fashion as $f_2 = \varphi \circ \mathfrak {u}$. Here $\varphi$ is the same, and $\mathfrak {u} = (\operatorname {mim}\langle \mathfrak {W}_1, \mathbb {P}_2\rangle , \ldots , \operatorname {mim}\langle \mathfrak {W}_n, \mathbb {P}_2\rangle )$, where the $\mathfrak {W}_i$ are arbitrary maximal chains embedded in the $\sigma$-algebra of the corresponding secrets of the conspiracy space $\Phi _2$. Since both $\mathbb {P}_1 \circ \mathfrak {r}^{-1}$ and $\mathbb {P}_2 \circ \mathfrak {u}^{-1}$ coincide with the Lebesgue measure, we conclude that $\mathbb {P}_1 \circ f_1^{-1} = \mathbb {P}_2 \circ f_2^{-1}$ as well, and this completes the proof of the theorem. $\quad \blacksquare$