A well-defined history in measurable factor spaces

Matthias G. Mayer

This is a technical post researching infinite factor spaces

In a factor space , for a measurable $Z : F \to Z (F)$ , a $Z$ -measurable index function $J : F \to {0, 1}^{I}$ is generating $X : F \to X (F)$ , if $X$ depends only on those arguments $(x_{i})$ , 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 $σ$ -algebra and $N_{i}$ is a $σ$ -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 $σ$ -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 $A * \subseteq B :\Leftrightarrow A ∖ B \in N$ .

Definition 2 (almost sure union with arbitrary index set)

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

To start, choose any $j \in J$ , $J_{0} = {j}$ and let $B_{0} = A_{j}$ .

Let $B_{n}$ be defined and let $s_{n} = sup j \in J P (A_{j} ∖ B_{n})$ . Choose $j_{n} \in J$ with $P ({A_{j}}_{n} ∖ B_{n}) \geq s_{n} - \frac{1}{n}$ and set $J_{n + 1} = J_{n} \cup {j_{n}}$ and $B_{n + 1} = B_{n} \cup {A_{j}}_{n}$ .

We set $J^{*} = {j_{n} : n \in N}$ and $A = ⋃_{j \in J^{*}} A_{j} .$

A contains $A_{j}$ : Let $j \in J$ , we have to show $P (A_{j} ∖ A) = 0$ . Assume not, then $P (A_{j} ∖ B_{n}) ↓ P (A_{j} ∖ A) = p > 0$ , so $s_{n} \geq p$ for all $n$ . Therefore, we have $P ({A_{j}}_{n} ∖ B_{n}) \geq p - \frac{1}{n}$ . Now $1 \geq P (A) = P (⨃_{n \in N} {A_{j}}_{n} ∖ B_{n}) = \sum_{n \in N} P ({A_{j}}_{n} ∖ B_{n}) = \infty$ , which is a contradiction.
Minimality and uniqueness: Now let $B \in A : \forall j \in J : A_{j} * \subseteq B$ . Then clearly, $A = ⋃_{n \in N} {A_{j}}_{n} * \subseteq B$ . If $B$ is also minimal, then clearly $A * \subseteq B * \subseteq A \Leftrightarrow P (A ▲ B) = 0 \Leftrightarrow A a.s. = B .$

Definition 3 (almost sure intersection)

We set ${⋂^{*}}_{j \in J} A_{j} = {({⋃^{*}}_{j \in J} A_{j}^{c})}_{j \in J}^{c}$ .

Definition 4 (feature)

We call a measurable function $Z : F \to Z (F)$ 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 $J_{i} = π_{i} \circ J$ . We identify $J$ with $J^{'} : F \to P (I)$ , where $J^{'} (x) = {i \in I : J_{i} (x) = 1}$ to allow for set operations such as $J \cap K = m i n (J, K)$ .

Definition 6

For an index function $J$ , we set $π_{J} = {(1_{{J_{i} = 1}} π_{i})}_{i \in I}$ .

Definition 7

We define that $X$ is almost surely a function of $Y$ by $X ≲^{*} Y :\Leftrightarrow σ (X) \subseteq σ (Y, N)$ .

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

Definition 8 (conditional generation)

Let $J : F \to {0, 1}^{I}$ be $Z -measurable$ . We write $J ⊢ X | Z$ , if $X ≲^{*} (π_{J}, Z) \land \forall P \in ▲^{*} (F, N) : π_{J} ⫫_{P} π_{¯ J} | Z$ . Note that $π_{J} ⫫_{P} π_{¯ J} ∣ ∣ Z \Leftrightarrow (π_{J}, Z) ⫫_{P} (π_{¯ J}, Z) ∣ ∣ Z$

Lemma 9

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

Then $σ (π_{J}, Z) = σ ({(π_{J^{n}}, Z)}_{n \in N})$ . Proof. Let $A ≔ σ (π_{J}, Z)$ , $B ≔ σ ({(π_{J^{n}}, Z)}_{n \in N})$ .

' $A \subseteq B$ ': Obviously, $σ (Z) \subseteq B$ . Let $A \in σ (1_{{J_{i} = 1}} π_{i})$ , then there is a $B \in σ (π_{i})$ with $\begin{matrix} A & = B \cap {J_{i} = 1} = B \cap ⋃ n \in N {J_{i}^{n} = 1} = ⋃ n \in N (B \cap {J_{i}^{n} = 1}) \in B . \end{matrix}$

' $A \supseteq B$ ': Obviously, $σ (Z) \subseteq A$ . Let $A \in σ (1_{{J_{i}^{n} = 1}} π_{i})$ , then there is $B \in σ (π_{i})$ with $\begin{matrix} A & = B \cap {J_{i}^{n} = 1} = B \cap {J_{i} = 1}      \in A \cap {J_{i}^{n} = 1}      \in σ (Z) \subseteq A \in A . \end{matrix}$

Lemma 10

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

Proof. Let $P \in ▲^{*} (F, N)$ . We first show $π_{J \cap K} ⫫_{P} π_{¯ J \cap K} | Z$ . For $i, j \in {0, 1}$ , let $M_{i j} = {J_{i}^{'} \cap K_{j}^{'}}$ where $J_{0}^{'} = J, J_{1}^{'} = ¯ J,$ etc. Let $A_{i j} \in M_{i j}$ . We have $({π_{M}}_{00}, {π_{M}}_{01}) ⫫_{P} ({π_{M}}_{10}, {π_{M}}_{11}) | Z, (1) ({π_{M}}_{00}, {π_{M}}_{10}) ⫫_{P} ({π_{M}}_{01}, {π_{M}}_{11}) | Z, (2)$ Now $\begin{matrix} P (⋂ i j A_{i j} | Z) & (1) = P (A_{00} \cap A_{01} | Z) \cdot P (A_{10} \cap A_{11} | Z) (2) = \prod i j P (A_{i j} | Z) . \end{matrix}$

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

It remains to show $X ≲^{*} (π_{J \cap K}, Z)$ . Since $π_{J \cap K} ⫫_{P} π_{J ∖ K} | Z$ , we have $P (π_{J ∖ K} | Z) = P (π_{J ∖ K} | π_{J \cap K}, Z)$ . Since the same holds true for $K ∖ J$ , we get $π_{J ∖ K} ⫫_{P} π_{K ∖ J} | π_{J \cap K}, Z$ and therefore $π_{J} ⫫_{P} π_{K} | π_{J \cap K}, Z$ .

We claim $σ (π_{J}, Z) \cap σ (π_{K}, Z) a.s. = σ (π_{J \cap K}, Z)$ .

' $\supseteq$ ': Trivial.
' $\subseteq^{*}$ ': Let $A \in σ (π_{J}, Z) \cap σ (π_{K}, Z)$ and $B = σ (π_{J \cap K}, Z)$ . Then $A ⫫_{P} A | B$ . Therefore, $P (A | B) = 1_{B}$ for a $B \in B$ . We claim $P (A ▲ C) = 0$ . We have $P (A ∖ C) = E (P (A ∖ C | B)) = E (1_{C^{c}} P (A ∖ C | B)) = E (0) = 0$ , we have $P (A ∖ C) = 0$ . Similarly, $P (C ∖ A | B) = 0$ and therefore $P (C ∖ A) = 0$ .

Lemma 11

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

We first show $π_{J} ⫫_{P} π_{¯ J} | Z$ . Since $σ (π_{J}) \subseteq σ (π_{K^{n}})$ , we have $π_{J} ⫫_{P} π_{^{n}} | Z$ . Now $^{n} ↑ ¯ J$ . Clearly, $⋃_{n \in N} σ (π_{^{n}})$ is $\cap$ -stable and by Lemma 9 a generator of $¯ J$ . Therefore, $π_{J} ⫫_{P} π_{¯ J} | Z$ .

We now show $X ≲^{*} (π_{J}, Z)$ . Let $B = ⋂_{n \in N} σ (π_{K^{n}}, Z)$ and $C = σ (π_{J}, Z)$

' $B \supseteq C$ " : Trivial.
' $B \subseteq^{*} C$ ': From $π_{J} ⫫_{P} π_{¯ J} | Z$ , we have $π_{J} ⫫_{P} π_{K^{n} ∖ J} | Z, π_{J} ⫫_{P} π_{^{n}} | Z$ Therefore, $π_{K^{n}} ⫫ π_{^{n} \cup J} | C$ .

Clearly, $⋃_{n \in N} σ (π_{^{n} \cup J})$ 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 $B \subseteq σ (C, N)$ .

Definition 12 (history)

$H (X | Y)$ 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 ${(H (X | Y))}_{i} ≔ 1_{⋂^{*} M_{i}}$ . Now clearly, for each $i$ , there are countable many $J^{n, i} ⊢ X | Z$ , such that ${(⋂_{n \in N} J^{n, i})}_{i}^{n, i} = 1_{⋂^{*} M_{i}}$ . Now $⋂_{n \in N, i \in I} J^{n_{i}} = H (X | Y)$ , and therefore $H (X | Y) ⊢ X | Z$ . Furthermore, by construction we have for any $J ⊢ X | Z$ , that ${{H (X | Y)}_{i} = 1} * \subseteq {J_{i} = 1}$ which implies $H (X | Y) * \subseteq J$ .

Uniqueness: Let $J$ be minimal, then clearly, $H (X | Y) * \subseteq J * \subseteq H (X | Y)$ .