# A well-defined history in measurable factor spaces

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This is a technical post researching infinite [factor spaces](https://www.lesswrong.com/posts/nksgLdLc7y5KtbmFw/a-reformulation-of-finite-factored-sets)

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

We will reintroduce all the definitions from the finite case:

Let $I$ be finite. Let
$\left( F_{i},\mathcal{A}_{i},\mathcal{N}_{i} \right)$ be a measure
space with nullsets, i.e. $F_{i}$ is a set, $\mathcal{A}_{i}$ is a
$\sigma$-algebra and $\mathcal{N}_{i}$ is a $\sigma$-ideal that admits a
probability, i.e. there is $P_{i}$ with
$\mathcal{N}_{i} = \left\{ A \in \mathcal{A}:P_{i}(A) = 0 \right\}$.

We define a product $\sigma$-ideal:
$\mathcal{N}_{1} \otimes \mathcal{N}_{2} = \left\{ A \in \mathcal{A}:\left\{ x:A(x) \notin \mathcal{N}_{2} \right\} \in \mathcal{N}_{1} \right\}.$
Clearly,
$\mathcal{N}_{1} \otimes \mathcal{N}_{2} = \left\{ A \in \mathcal{A}:P_{1} \times P_{2}(A) = 0 \right\}$.
We can extend this by induction to
$\bigotimes_{i \in I}\mathcal{N}_{i}$.

We construct
$\left( F,\mathcal{A},\mathcal{N} \right) = \left( ⨉_{i \in I}F,\bigotimes_{i \in I}\mathcal{A}_{i},\bigotimes_{i \in I}\mathcal{N}_{i} \right)$.
Furthermore, let $P = ⨉_{i \in I}P_{i}$.

*Definition* 1 (almost sure subset)

   We will write
$A\overset{\ast}{\subseteq}B: \Leftrightarrow A \smallsetminus B \in \mathcal{N}$.
 

*Definition* 2 (almost sure union with arbitrary index set)

   If $\mathcal{N}$ is a $\sigma$-ideal that admits a probability $P$ and
$\left( A_{j} \right)_{j \in J} \in \mathcal{A}^{J}$ is a family of
measurable sets, then there exists an almost surely unique and minimal
$A \in \mathcal{A}$ with
$$\forall j \in J:A_{j}\overset{\ast}{\subseteq}A.$$ i.e. whenever
$\forall j \in J:A_{j}\overset{\ast}{\subseteq}B$, we have
$A\overset{\ast}{\subseteq}B$. 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
$s_{n} = \sup\limits_{j \in J}P\left( A_{j} \smallsetminus B_{n} \right)$.
Choose $j_{n} \in J$ with
$P\left( {A_{j}}_{n} \smallsetminus B_{n} \right) \geq s_{n} - \frac{1}{n}$
and set $J_{n + 1} = J_{n} \cup \left\{ j_{n} \right\}$ and
$B_{n + 1} = B_{n} \cup {A_{j}}_{n}$.

We set $J^{\ast} = \left\{ j_{n}:n \in {\mathbb{N}} \right\}$ and
$A = \bigcup_{j \in J^{\ast}}A_{j}.$

-   A contains $A_{j}$: Let $j \in J$, we have to show
    $P\left( A_{j} \smallsetminus A \right) = 0$. Assume not, then
    $P\left( A_{j} \smallsetminus B_{n} \right) \downarrow P\left( A_{j} \smallsetminus A \right) = p > 0$,
    so $s_{n} \geq p$ for all $n$. Therefore, we have
    $P\left( {A_{j}}_{n} \smallsetminus B_{n} \right) \geq p - \frac{1}{n}$.
    Now
    $1 \geq P(A) = P\left( ⨃_{n \in {\mathbb{N}}}{A_{j}}_{n} \smallsetminus B_{n} \right) = \sum_{n \in {\mathbb{N}}}P\left( {A_{j}}_{n} \smallsetminus B_{n} \right) = \infty$,
    which is a contradiction.

-   Minimality and uniqueness: Now let
    $B \in \mathcal{A}:\forall j \in J:A_{j}\overset{\ast}{\subseteq}B$.
    Then clearly,
    $A = \bigcup_{n \in {\mathbb{N}}}{A_{j}}_{n}\overset{\ast}{\subseteq}B$.
    If $B$ is also minimal, then clearly
    $$A\overset{\ast}{\subseteq}B\overset{\ast}{\subseteq}A \Leftrightarrow P(A▲B) = 0 \Leftrightarrow A\overset{\text{a.s. }}{=}B.$$

 

*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
$\mathcal{Z}:F \rightarrow \mathcal{Z}(F)$ a feature.   In the
following let $\mathcal{X},\mathcal{Y},\mathcal{Z}$ be features.

*Definition* 5 (index function)

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

*Definition* 6

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

*Definition* 7

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

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

*Definition* 8 (conditional generation)

   Let $J:F \rightarrow \left\{ 0,1 \right\}^{I}$ be
$\mathcal{Z}\text{-measurable}$. We write
$J \vdash \mathcal{X}~|~\mathcal{Z}$, if
$\mathcal{X} \lesssim^{\ast}\left( \pi_{J},\mathcal{Z} \right) \land \forall P \in ▲^{\ast}(F,\mathcal{N}):\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$. Note that $\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}\left| \mathcal{Z} \Leftrightarrow \left( \pi_{J},\mathcal{Z} \right){
 ⫫ {}}_{P}\left( \pi_{\overset{¯}{J}},\mathcal{Z} \right) \right|\mathcal{Z}$
 

*Lemma* 9

For $n \in {\mathbb{N}}$ let $J^{n}:F \rightarrow {\mathcal{P}(I)}$ be
$\mathcal{Z}\text{-measurable}$ and
$J = \bigcup_{n \in {\mathbb{N}}}J^{n}$.

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

'$\mathcal{A} \subseteq \mathcal{B}$': Obviously,
$\sigma(\mathcal{Z}) \subseteq \mathcal{B}$. Let
$A \in \sigma(1_{\left\{ J_{i} = 1 \right\}}\pi_{i})$, then there is a
$B \in \sigma(\pi_{i})$ with $$\begin{aligned}
A & = B \cap \left\{ J_{i} = 1 \right\} \\
 & = B \cap \bigcup_{n \in {\mathbb{N}}}\left\{ J_{i}^{n} = 1 \right\} \\
 & = \bigcup_{n \in {\mathbb{N}}}\left( B \cap \left\{ J_{i}^{n} = 1 \right\} \right) \in \mathcal{B}.
\end{aligned}$$

'$\mathcal{A} \supseteq \mathcal{B}$': Obviously,
$\sigma(\mathcal{Z}) \subseteq \mathcal{A}$. Let
$A \in \sigma(1_{\left\{ J_{i}^{n} = 1 \right\}}\pi_{i})$, then there is
$B \in \sigma(\pi_{i})$ with $$\begin{aligned}
A & = B \cap \left\{ J_{i}^{n} = 1 \right\} \\
 & = \underset{\in \mathcal{A}}{\underbrace{B \cap \left\{ J_{i} = 1 \right\}}} \cap \underset{\in \sigma(\mathcal{Z}) \subseteq \mathcal{A}}{\underbrace{\left\{ J_{i}^{n} = 1 \right\}}} \in \mathcal{A}.
\end{aligned}$$  

*Lemma* 10

Let $J,K:F \rightarrow {\mathcal{P}(I)}$ and
$J,K \vdash \mathcal{X}~|~\mathcal{Z}$. Then
$J \cap K \vdash \mathcal{X}~|~\mathcal{Z}$.  

*Proof.*    Let $P \in ▲^{\ast}(F,\mathcal{N})$. We first show
$\pi_{J \cap K}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J \cap K}}~|~\mathcal{Z}$. For
$i,j \in \left\{ 0,1 \right\}$, let
$M_{ij} = \left\{ J'_{i} \cap K'_{j} \right\}$ where
$$J'_{0} = J,J'_{1} = \overset{¯}{J},$$ etc. Let
$A_{ij} \in M_{ij}$. We have
$$\left( {\pi_{M}}_{00},{\pi_{M}}_{01} \right){
 ⫫ {}}_{P}\left( {\pi_{M}}_{10},{\pi_{M}}_{11} \right)\left.
|\mathcal{Z},\quad\ (1)\  \\
\left( {\pi_{M}}_{00},{\pi_{M}}_{10} \right){
 ⫫ {}}_{P}\left( {\pi_{M}}_{01},{\pi_{M}}_{11} \right)|
\right.\mathcal{Z},\quad\ (2)\ $$ Now $$\begin{aligned}
P\left( \bigcap_{ij}A_{ij}~|~\mathcal{Z} \right) & \overset{(1)\ }{=}P\left( A_{00} \cap A_{01}~|~\mathcal{Z} \right) \cdot P\left( A_{10} \cap A_{11}~|~\mathcal{Z} \right) \\
 & \overset{(2)\ }{=}\prod_{ij}P\left( A_{ij}~|~\mathcal{Z} \right).
\end{aligned}$$

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

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

We claim
$\sigma(\pi_{J},\mathcal{Z}) \cap \sigma(\pi_{K},\mathcal{Z})\overset{\text{a.s. }}{=}\sigma(\pi_{J \cap K},\mathcal{Z})$.

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

*Lemma* 11

For $n \in {\mathbb{N}}$, let $J^{n} \vdash \mathcal{X}~|~\mathcal{Z}$.
Let $J = \bigcap_{n \in {\mathbb{N}}}J^{n}$. Then
$J \vdash \mathcal{X}~|~\mathcal{Z}$.  
*Proof.*   Let $K^{n} = \bigcap_{k \leq n}J^{k}$. Clearly,
$K^{n} \vdash \mathcal{X}~|~\mathcal{Z}$ and $K^{n} \downarrow J$.

We first show $\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$. Since
$\sigma(\pi_{J}) \subseteq \sigma(\pi_{K^{n}})$, we have $\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{K^{n}}}~|~\mathcal{Z}$. Now
$\overset{¯}{K^{n}} \uparrow \overset{¯}{J}$. Clearly,
$\bigcup_{n \in {\mathbb{N}}}\sigma(\pi_{\overset{¯}{K^{n}}})$ is
$\cap$-stable and by Lemma 9 a generator of $\overset{¯}{J}$. Therefore,
$\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$.

We now show
$\mathcal{X} \lesssim^{\ast}\left( \pi_{J},\mathcal{Z} \right)$. Let
$\mathcal{B} = \bigcap_{n \in {\mathbb{N}}}\sigma(\pi_{K^{n}},\mathcal{Z})$
and $\mathcal{C} = \sigma(\pi_{J},\mathcal{Z})$

'$\mathcal{B} \supseteq \mathcal{C}$" : Trivial.\
'$\mathcal{B} \subseteq^{\ast}\mathcal{C}$': From $\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$, we have
$$
\pi_{J} {
 ⫫ {}}_{P}\pi_{K^{n} \smallsetminus J}\left.
|\mathcal{Z}, \\
\pi_{J} {
 ⫫ {}}_{P}\pi_{\overset{¯}{K^{n}}}|
\right.\mathcal{Z}
$$ Therefore, $\pi_{K^{n}}{
 ⫫ {}}\pi_{\overset{¯}{K^{n}} \cup J}~|~\mathcal{C}$.

Clearly,
$\bigcup_{n \in {\mathbb{N}}}\sigma(\pi_{\overset{¯}{K^{n}} \cup J})$
is a $\cap$-stable generator of $\mathcal{A}$ and therefore
$\mathcal{B}{
 ⫫ {}}_{P}\mathcal{A}~|~\mathcal{C}$. Now, since
$\mathcal{B} \subseteq \mathcal{A}$, we have $\mathcal{B}{
 ⫫ {}}_{P}\mathcal{B}~|~\mathcal{C}$ and therefore
$\mathcal{B} \subseteq \sigma(\mathcal{C},\mathcal{N})$.  

*Definition* 12 (history)

   $\mathcal{H}(\mathcal{X}|\mathcal{Y})$ is the a.s. smallest
generating index function.  

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

*Proof.* Existence: Let
$M_{i} = \left\{ \left\{ J_{i} = 1 \right\}:J \vdash \mathcal{X}~|~\mathcal{Z} \right\}$.
Define
$\left( \mathcal{H}\left( \mathcal{X}|\mathcal{Y} \right) \right)_{i} ≔ 1_{\bigcap^{\ast}M_{i}}$.
Now clearly, for each $i$, there are countable many
$J^{n,i} \vdash \mathcal{X}~|~\mathcal{Z}$, such that
$\left( \bigcap_{n \in {\mathbb{N}}}J^{n,i} \right)_{i} = 1_{\bigcap^{\ast}M_{i}}$.
Now
$\bigcap_{n \in {\mathbb{N}},i \in I}J^{n_{i}} = \mathcal{H}(\mathcal{X}|\mathcal{Y})$,
and therefore
$\mathcal{H}(\mathcal{X}|\mathcal{Y}) \vdash \mathcal{X}~|~\mathcal{Z}$.
Furthermore, by construction we have for any
$J \vdash \mathcal{X}~|~\mathcal{Z}$, that
$\left\{ {\mathcal{H}(\mathcal{X}|\mathcal{Y})}_{i} = 1 \right\}\overset{\ast}{\subseteq}\left\{ J_{i} = 1 \right\}$
which implies
$\mathcal{H}(\mathcal{X}|\mathcal{Y})\overset{\ast}{\subseteq}J$.

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