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This is a technical post researching infinite factor spaces

In a factor space $F={⨉}_{i\in I}{F}_i$, for a measurable $Z:F\to Z\left(F\right)$, a $Z$-measurable index function $J:F\to {\{0,1\}}^I$ is generating $X:F\to X\left(F\right)$, if $X$ depends only on those arguments $\left(x_i\right)$, where $J(x{)}_i$ is $1$, and $Z$. In this post, we show that there is an almost surely minimal such index function that we call the history of $X$ given $Z$.

We will reintroduce all the definitions from the finite case:

Let $I$ be finite. Let $(F_i,A_i,N_i)$ be a measure space with nullsets, i.e. $F_i$ is a set, $A_i$ is a $\sigma$-algebra and $N_i$ is a $\sigma$-ideal that admits a probability, i.e. there is $P_i$ with $N_i=\{A\in A:P_i(A)=0\}$.

We define a product $\sigma$-ideal: $N_1\otimes N_2=\{A\in A:\{x:A(x)\notin N_2\}\in N_1\}.$ Clearly, $N_1\otimes N_2=\{A\in A:P_1\times P_2(A)=0\}$. We can extend this by induction to ${⨂}_{i\in I}{N}_i$.

We construct $(F,A,N)=({⨉}_{i\in I}F,{⨂}_{i\in I}{A}_i,{⨂}_{i\in I}{N}_i)$. Furthermore, let $P={⨉}_{i\in I}{P}_i$.

Definition 1 (almost sure subset)

We will write $A\stackrel{\ast }{\subseteq }B:\iff A\setminus B\in N$.

Definition 2 (almost sure union with arbitrary index set)

If $N$ is a $\sigma$-ideal that admits a probability $P$ and ${\left(A_j\right)}_{j\in J}\in A^J$ is a family of measurable sets, then there exists an almost surely unique and minimal $A\in A$ with $$\forall j\in J:A_j\stackrel{\ast }{\subseteq }A.$$ i.e. whenever $\forall j\in J:A_j\stackrel{\ast }{\subseteq }B$, we have $A\stackrel{\ast }{\subseteq }B$. We set $${\bigcup ^{\ast }}_{j\in J}{A}_j≔A.$$ Furthermore, there is $J^{\ast }\subseteq J$ countable, s.t. ${{\bigcup }^{\ast }}_{j\in J}{A}_j={\bigcup }_{j\in J^{\ast }}{A}_j$.
Proof. We will construct $J^{\ast }$.

To start, choose any $j\in J$, $J_0=\left\{j\right\}$ and let $B_0=A_j$.

Let $B_n$ be defined and let $s_n=\underset{j\in J}{sup}P(A_j\setminus B_n)$. Choose $j_n\in J$ with $P({A_j}_n\setminus B_n)\ge s_n-\frac{1}{n}$ and set $J_{n+1}=J_n\cup \left\{j_n\right\}$ and $B_{n+1}=B_n\cup {A_j}_n$.

We set $J^{\ast }=\{j_n:n\in N\}$ and $A={\bigcup }_{j\in J^{\ast }}{A}_j.$

• A contains $A_j$: Let $j\in J$, we have to show $P(A_j\setminus A)=0$. Assume not, then $P(A_j\setminus B_n)\downarrow P(A_j\setminus A)=p>0$, so $s_n\ge p$ for all $n$. Therefore, we have $P({A_j}_n\setminus B_n)\ge p-\frac{1}{n}$. Now $1\ge P\left(A\right)=P({⨃}_{n\in N}{A_j}_n\setminus B_n)={\sum }_{n\in N}P({A_j}_n\setminus B_n)=\infty$, which is a contradiction.

• Minimality and uniqueness: Now let $B\in A:\forall j\in J:A_j\stackrel{\ast }{\subseteq }B$. Then clearly, $A={\bigcup }_{n\in N}{A_j}_n\stackrel{\ast }{\subseteq }B$. If $B$ is also minimal, then clearly $$A\stackrel{\ast }{\subseteq }B\stackrel{\ast }{\subseteq }A\iff P(A▲B)=0\iff A\stackrel{\text{a.s.}}{=}B.$$

Definition 3 (almost sure intersection)

We set ${{\bigcap }^{\ast }}_{j\in J}{A}_j={\left({{\bigcup }^{\ast }}_{j\in J}{A}_j^c\right)}^c$.

Definition 4 (feature)

We call a measurable function $Z:F\to Z\left(F\right)$ a feature. In the following let $X,Y,Z$ be features.

Definition 5 (index function)

We call a measurable $J:F\to {\{0,1\}}^I$ an index function. We write $J_i={\pi }_i\circ J$. We identify $J$ with $J^{\prime }:F\to P\left(I\right)$, where $J^{\prime }\left(x\right)=\{i\in I:J_i(x)=1\}$ to allow for set operations such as $J\cap K=min(J,K)$.

Definition 6

For an index function $J$, we set ${\pi }_J={\left(1_{\{J_i=1\}}{\pi }_i\right)}_{i\in I}$.

Definition 7

We define that $X$ is almost surely a function of $Y$ by $X{\lesssim }^{\ast }Y:\iff \sigma \left(X\right)\subseteq \sigma (Y,N)$.

Let ${▲}^{\ast }(F,N)$ be all product distributions whose nullsets are $N$.

Definition 8 (conditional generation)

Let $J:F\to {\{0,1\}}^I$ be $Z\text{-measurable}$. We write $J⊢X|Z$, if $X{\lesssim }^{\ast }({\pi }_J,Z)\wedge \forall P\in {▲}^{\ast }(F,N):{\pi }_J{⫫}_P{\pi }_{\overline{J}}|Z$. Note that ${\pi }_J{⫫}_P{\pi }_{\overline{J}}|Z\iff ({\pi }_J,Z){⫫}_P({\pi }_{\overline{J}},Z)|Z$

Lemma 9

For $n\in N$ let $J^n:F\to P\left(I\right)$ be $Z\text{-measurable}$ and $J={\bigcup }_{n\in N}{J}^n$.

Then $\sigma ({\pi }_J,Z)=\sigma \left({({\pi }_{J^n},Z)}_{n\in N}\right)$. Proof. Let $A≔\sigma ({\pi }_J,Z)$, $B≔\sigma \left({({\pi }_{J^n},Z)}_{n\in N}\right)$.

'$A\subseteq B$': Obviously, $\sigma \left(Z\right)\subseteq B$. Let $A\in \sigma \left(1_{\{J_i=1\}}{\pi }_i\right)$, then there is a $B\in \sigma \left({\pi }_i\right)$ with $$\begin{array}{cc}A& =B\cap \{J_i=1\}\\ & =B\cap \bigcup _{n\in N}\{J_i^n=1\}\\ & =\bigcup _{n\in N}(B\cap \{J_i^n=1\})\in B.\end{array}$$

'$A\supseteq B$': Obviously, $\sigma \left(Z\right)\subseteq A$. Let $A\in \sigma \left(1_{\{J_i^n=1\}}{\pi }_i\right)$, then there is $B\in \sigma \left({\pi }_i\right)$ with $$\begin{array}{cc}A& =B\cap \{J_i^n=1\}\\ & =\underset{\in A}{\underset{}{B\cap \{J_i=1\}}}\cap \underset{\in \sigma \left(Z\right)\subseteq A}{\underset{}{\{J_i^n=1\}}}\in A.\end{array}$$

Lemma 10

Let $J,K:F\to P\left(I\right)$ and $J,K⊢X|Z$. Then $J\cap K⊢X|Z$.

Proof. Let $P\in {▲}^{\ast }(F,N)$. We first show ${\pi }_{J\cap K}{⫫}_P{\pi }_{\overline{J\cap K}}|Z$. For $i,j\in \{0,1\}$, let $M_{ij}=\{J_i^{\prime }\cap K_j^{\prime }\}$ where $$J_0^{\prime }=J,J_1^{\prime }=\overline{J},$$ etc. Let $A_{ij}\in M_{ij}$. We have $$({{\pi }_M}_{00},{{\pi }_M}_{01}){⫫}_P({{\pi }_M}_{10},{{\pi }_M}_{11})|Z,\left(1\right)({{\pi }_M}_{00},{{\pi }_M}_{10}){⫫}_P({{\pi }_M}_{01},{{\pi }_M}_{11})|Z,\left(2\right)$$ Now $$\begin{array}{cc}P\left(\bigcap _{ij}{A}_{ij}|Z\right)& \stackrel{\left(1\right)}{=}P(A_{00}\cap A_{01}|Z)\cdot P(A_{10}\cap A_{11}|Z)\\ & \stackrel{\left(2\right)}{=}\prod _{ij}P\left(A_{ij}|Z\right).\end{array}$$

Now since sets of the form $A_{01}\cap A_{10}\cap A_{11}$ generate ${\pi }_{\overline{J\cap K}}$ and are $\cap$-stable, we have that ${\pi }_{J\cap K}{⫫}_P{\pi }_{\overline{J\cap K}}|Z$.

It remains to show $X{\lesssim }^{\ast }({\pi }_{J\cap K},Z)$. Since ${\pi }_{J\cap K}{⫫}_P{\pi }_{J\setminus K}|Z$, we have $P\left({\pi }_{J\setminus K}|Z\right)=P({\pi }_{J\setminus K}|{\pi }_{J\cap K},Z)$. Since the same holds true for $K\setminus J$, we get ${\pi }_{J\setminus K}{⫫}_P{\pi }_{K\setminus J}|{\pi }_{J\cap K},Z$ and therefore ${\pi }_J{⫫}_P{\pi }_K|{\pi }_{J\cap K},Z$.

We claim $\sigma ({\pi }_J,Z)\cap \sigma ({\pi }_K,Z)\stackrel{\text{a.s.}}{=}\sigma ({\pi }_{J\cap K},Z)$.

'$\supseteq$': Trivial.
'${\subseteq }^{\ast }$': Let $A\in \sigma ({\pi }_J,Z)\cap \sigma ({\pi }_K,Z)$ and $B=\sigma ({\pi }_{J\cap K},Z)$. Then $A{⫫}_{P}A|B$. Therefore, $P\left(A|B\right)=1_B$ for a $B\in B$. We claim $P(A▲C)=0$. We have $P(A\setminus C)=E\left(P(A\setminus C|B)\right)=E\left(1_{C^c}P(A\setminus C|B)\right)=E\left(0\right)=0$, we have $P(A\setminus C)=0$. Similarly, $P(C\setminus A|B)=0$ and therefore $P(C\setminus A)=0$.

Lemma 11

For $n\in N$, let $J^n⊢X|Z$. Let $J={\bigcap }_{n\in N}{J}^n$. Then $J⊢X|Z$.
Proof. Let $K^n={\bigcap }_{k\le n}{J}^k$. Clearly, $K^n⊢X|Z$ and $K^n\downarrow J$.

We first show ${\pi }_J{⫫}_P{\pi }_{\overline{J}}|Z$. Since $\sigma \left({\pi }_J\right)\subseteq \sigma \left({\pi }_{K^n}\right)$, we have ${\pi }_J{⫫}_P{\pi }_{\overline{K^n}}|Z$. Now $\overline{K^n}\uparrow \overline{J}$. Clearly, ${\bigcup }_{n\in N}\sigma \left({\pi }_{\overline{K^n}}\right)$ is $\cap$-stable and by Lemma 9 a generator of $\overline{J}$. Therefore, ${\pi }_J{⫫}_P{\pi }_{\overline{J}}|Z$.

We now show $X{\lesssim }^{\ast }({\pi }_J,Z)$. Let $B={\bigcap }_{n\in N}\sigma ({\pi }_{K^n},Z)$ and $C=\sigma ({\pi }_J,Z)$

'$B\supseteq C$" : Trivial.
'$B{\subseteq }^{\ast }C$': From ${\pi }_J{⫫}_P{\pi }_{\overline{J}}|Z$, we have $${\pi }_J{⫫}_P{\pi }_{K^n\setminus J}|Z,{\pi }_J{⫫}_P{\pi }_{\overline{K^n}}|Z$$ Therefore, ${\pi }_{K^n}⫫{\pi }_{\overline{K^n}\cup J}|C$.

Clearly, ${\bigcup }_{n\in N}\sigma \left({\pi }_{\overline{K^n}\cup J}\right)$ is a $\cap$-stable generator of $A$ and therefore $B{⫫}_{P}A|C$. Now, since $B\subseteq A$, we have $B{⫫}_{P}B|C$ and therefore $B\subseteq \sigma (C,N)$.

Definition 12 (history)

$H\left(X|Y\right)$ is the a.s. smallest generating index function.

Theorem 13 A history exists and is a.s. unique.

Proof. Existence: Let $M_i=\{\{J_i=1\}:J⊢X|Z\}$. Define ${\left(H\left(X|Y\right)\right)}_i≔1_{{\bigcap }^{\ast }{M}_i}$. Now clearly, for each $i$, there are countable many $J^{n,i}⊢X|Z$, such that ${\left({\bigcap }_{n\in N}{J}^{n,i}\right)}_i=1_{{\bigcap }^{\ast }{M}_i}$. Now ${\bigcap }_{n\in N,i\in I}{J}^{n_i}=H\left(X|Y\right)$, and therefore $H\left(X|Y\right)⊢X|Z$. Furthermore, by construction we have for any $J⊢X|Z$, that $\{{H\left(X|Y\right)}_i=1\}\stackrel{\ast }{\subseteq }\{J_i=1\}$ which implies $H\left(X|Y\right)\stackrel{\ast }{\subseteq }J$.

Uniqueness: Let $J$ be minimal, then clearly, $H\left(X|Y\right)\stackrel{\ast }{\subseteq }J\stackrel{\ast }{\subseteq }H\left(X|Y\right)$.