# A counterexample for 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)

I define a general measurable factor space that
shows that the history can't be straightforwardly generalized to the infinite case.
The example shows the following:

-   Unlike in the finite setting, the condition
     $\forall P \in ▲^{\ast}(F):\pi_{J}{
      ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$ in the definition of
     generation (see below) can't be characterized by requiring that the atoms of
     $\mathcal{Z}$ are rectangles. In the example, we show that this would
     imply an orthogonality where the implication $\mathcal{X}{
      \perp {}}\mathcal{Y}\left| \mathcal{Z} \Rightarrow \forall P \in ▲^{\ast}(F):\mathcal{X}{
      ⫫ {}}_{P}\mathcal{Y} \right|\mathcal{Z}$ is not true.

-  In measurable spaces, conditional histories can only be well-defined
    up to a chosen system of nullsets. In the example, we show that a
    minimal generating index function $J:F \rightarrow {\mathcal{P}(I)}$
    does not exist. (We will show in a later post that the history exists if we fix nullsets.)

To talk about orthogonality, we first have to introduce the concepts for
measurable spaces. For now, we focus on the index set $I$ being finite. A
straightforward generalization from the finite case is the following.
We have measurable spaces $\left( F_{i},\mathcal{A}_{i} \right)$ and
construct $F=\mathop{\Large\times\normalsize}_{i \in I} F_i$. Let $\mathcal{Z}:F \rightarrow \mathcal{Z}(F)$ be a feature
(i.e. measurable). Let $J:F \rightarrow {\mathcal{P}(I)}$ be
$\mathcal{Z}\text{-measurable}$. We define
$$\pi_{J}(x) = \pi_{J(x)}(x)$$ We note, that we assume w.l.o.g. that we
assume that $F_{i}$ are pairwise disjoint. Recall that $▲^{\ast}(F)$ is
the set of product probability distributions on $F$.

$$J \vdash \mathcal{X}\left| \mathcal{Z}: \Leftrightarrow \mathcal{X} \lesssim (\pi_{J},\mathcal{Z}) \land \forall P \in ▲^{\ast}(F):\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}} \right|\mathcal{Z}.$$

The history should now be the smallest such $J$. Furthermore, for
orthogonality defined by $\mathcal{X}{
 \perp {}}\mathcal{Y}~|~\mathcal{Z}: \Leftrightarrow \mathcal{H}(\mathcal{X}|\mathcal{Z}) \cap \mathcal{H}(\mathcal{Y}|\mathcal{Z}) = \varnothing$
we should have $\mathcal{X}{
 \perp {}}\mathcal{Y}~|~\mathcal{Z} \Leftrightarrow \forall P \in ▲^{\ast}(F):\mathcal{X}{
      ⫫ {}}_{P}\mathcal{Y} |\mathcal{Z}$

The idea for the example consists of choosing $\mathcal{Z}$, such that
$\mathcal{Z}^{- 1}(z)$ consists of four points, where the points will
always fall into one unit square each. Call those unit squares $U_{ij}$.
Choosing $\mathcal{Z}|_{U_{ij}}$ to be a diffeomorphism allows us to
calculate the conditional probabilities explicitly. Almost all choices
of $\mathcal{Z}$ where $\mathcal{Z}^{- 1}(z)$ is a rectangle for all
$z$, will fail to satisfy $\forall P \in ▲^{\ast}(F):\mathcal{X}{
 ⫫ {}}_{P}\mathcal{Y}~|~\mathcal{Z}$.

# Example 1

Let $I = \left\{ 1,2 \right\}$. For $i \in I$, let
$F_{i} = \lbrack 0,1\rbrack \cup \lbrack 2,3\rbrack$ endowed with the
Borel sigma-algebra.

Now
$F = F_{1} \times F_{2} = \left( \lbrack 0,1\rbrack \cup \lbrack 2,3\rbrack \right)\hat{}2$.

$F$ consists of 4 unit squares.

![](https://i.imgur.com/Y9tryrQ.png)


We will now define a $Z$, such that $Z^{- 1}$ is a rectangle of four
points.

Let $$U_{ij} = \begin{cases}
\lbrack 0,1\rbrack & \text{ if }i = 1 \\
\lbrack 2,3\rbrack & \text{else}
\end{cases} \times \begin{cases}
\lbrack 0,1\rbrack & \text{ if }j = 1 \\
\lbrack 2,3\rbrack & \text{else}
\end{cases}$$ be the four unit squares that $F$ consists of.

Let
$\varphi_{x},\varphi_{y}:\lbrack 0,1\rbrack^{2} \rightarrow \lbrack 2,3\rbrack$.
$\varphi_{x}$ will be the function that chooses the new $x$ coordinate.

Let $$\begin{aligned}
\pi_{x}:\lbrack 0,1\rbrack^{2} & \rightarrow \lbrack 2,3\rbrack, \\
x,y & \left. \begin{aligned}
| \rightarrow x + 2 \\
\pi_{y}:\lbrack 0,1\rbrack^{2} & \rightarrow \lbrack 2,3\rbrack, \\
x,y & |
\end{aligned} \right. \rightarrow y + 2
\end{aligned}$$

Let $$\begin{array}{r}
\varphi_{11} = \left( \pi_{x},\pi_{y} \right) \\
\varphi_{21} = \left( \varphi_{x},\pi_{y} \right) \\
\varphi_{12} = \left( \pi_{x},\varphi_{y} \right) \\
\varphi_{22} = \left( \varphi_{x},\varphi_{y} \right)
\end{array}$$

We assume that all $\varphi_{ij}$ are diffeomorphisms.

![](https://i.imgur.com/q0AvuCS.png)

Define $\mathcal{Z}_{ij} = \varphi_{ij}^{- 1}$ and $\mathcal{Z}$ by
$\mathcal{Z}|_{U_{ij}} = \mathcal{Z}_{ij}$.

First we assure ourselves, that $\mathcal{Z}$ indeed has rectangle
atoms.

Let $z = \left( z_{1},z_{2} \right) \in \lbrack 0,1\rbrack^{2}$, then
$$\begin{aligned}
\mathcal{Z}^{- 1}(z) = \left. \begin{aligned}
\{ & \varphi_{ij}(z)\}
\end{aligned} \right. \\
 = \left. \begin{aligned}
\{ & \left. \begin{array}{rlr}
(z_{1}, & & z_{2})
\end{array} \right., \\
 & \left. \begin{array}{rlr}
(z_{1}, & & \varphi_{y}\left( z_{2} \right))
\end{array} \right. \\
 & \left. \begin{array}{rlr}
(\varphi_{x}\left( z_{1} \right), & & \varphi_{y}\left( z_{2} \right))
\end{array} \right. \\
 & \left. \begin{array}{rlr}
(\varphi_{x}\left( z_{1} \right), & & z_{2})
\end{array} \right.\}
\end{aligned} \right. \\
 = & \left\{ z_{1},z_{2} \right\} & & \times \left\{ \varphi_{x}\left( z_{1} \right),\varphi_{y}\left( z_{2} \right) \right\}
\end{aligned}$$

Now the projections $\pi_{i}:F \rightarrow F_{i}$ would be orthogonal,
if we had $$\begin{array}{r}
\forall P \in ▲^{\ast}(F):\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z} \\
 \Leftrightarrow \forall z \in \mathcal{Z}(F):\mathcal{Z}^{- 1}(z)\text{ is a rectangle}.
\end{array}$$

We will now calculate the conditional probabilities by Lebesgue
differentiation.

We first note that $\mathcal{Z}$ is continuous. Each
$\mathcal{Z}^{- 1}(z)$ consists of four points, one in each $U_{ij}$
that form a rectangle.

Because $F$ is compact, it is easy to show that
$\mathcal{Z}^{- 1}\left( B_{\delta}(z) \right)$ concentrates around
$\mathcal{Z}^{- 1}(z)$ for small delta, i.e.

$$\forall\varepsilon > 0:\exists\delta > 0:d\left( \mathcal{Z}^{- 1}\left( B_{\delta}(z) \right),\mathcal{Z}^{- 1}(z) \right) < \varepsilon,$$
where $d(A,B) = \sup\limits_{a \in A}\inf\limits_{b \in B}d(a,b)$ is the
Hausdorff metric for the standard metric $d = || \cdot - \cdot ||_{2}$.

We will use the uniform distribution
$P = P_{1} \times P_{2} = \frac{\lambda{|_{F}}_{1}}{\lambda\left( F_{1} \right)} \times \frac{\lambda{|_{F}}_{2}}{\lambda\left( F_{2} \right)} = \frac{\lambda^{2}|_{F}}{\lambda^{2}(F)}$

to show $\pi_{1}{
 ⫫ {}}_{P}\pi_{2}|\mathcal{Z}$.

Let
$A = A_{1} \times A_{2} = \left( \left\lbrack 0,a_{1} \right\rbrack \cap F_{1} \right) \times \left( \left\lbrack 0,a_{2} \right\rbrack \cap F_{2} \right)$
for $a_{i} \in F_{i} \smallsetminus \left\{ 0 \right\}$.

Take a $z$ such that $\mathcal{Z}^{- 1}$ does not lie on the edge of
$A$, i.e.
$\mathcal{Z}^{- 1}(z) \cap F_{1} \times \left\{ a_{2} \right\} \cup \left\{ a_{1} \right\} \times F_{1} = \varnothing$

By Lebesgue differentiation, we know that
$$\frac{1}{B_{\delta}(z)}\int_{B_{\delta}(z)}P\left( A|\mathcal{Z} = \cdot \right)dP_{\mathcal{Z}}\overset{\delta \rightarrow 0}{\rightarrow}P\left( A|\mathcal{Z} = z \right)\text{ for  }P_{\mathcal{Z}}\text{-a.a z }$$
Since the left side equals
$P\left( A|\left\{ \mathcal{Z} \in B_{\delta}(z) \right\} \right)$, we
have
$z \mapsto \lim\limits_{\delta \rightarrow 0}P\left( A~|~\left\{ \mathcal{Z} \in B_{\delta}(z) \right\} \right)$
is a version of $P\left( A|\mathcal{Z} = \cdot \right)$.

Now we will split up $\left\{ \mathcal{Z} \in B_{\delta}(z) \right\}$
into its parts in the unit squares $U_{ij}$.

Let
$C_{ij\delta} = \left\{ \mathcal{Z}_{ij} \in B_{\delta}(z) \right\} = \varphi_{ij}\left( C_{11\delta} \right)$.
Let $c = \frac{1}{\lambda^{2}(F)}$.

We have $$\begin{aligned}
P\left( C_{ij\delta} \right) & = c\int_{\varphi_{ij}\left( C_{11\delta} \right)}1d\lambda^{2} \\
 & = c{\int_{C}}_{11\delta}1 \circ \varphi_{ij}\left| {\det D\varphi_{ij}} \right|d\lambda^{2} \\
 & = c{\int_{C}}_{11\delta}\left| {\det D\varphi_{ij}} \right|d\lambda^{2}
\end{aligned}$$

Since $\left| \det D\varphi_{ij} \right|$ is continuous, we have
$$\begin{aligned}
 & \frac{1}{P\left( C_{11\delta} \right)}{\int_{C}}_{11\delta}\left| {\det D\varphi_{ij}} \right|dP \\
 = & \frac{1}{\lambda^{2}\left( C_{11\delta} \right)}{\int_{C}}_{11\delta}\left| {\det D\varphi_{ij}} \right|d\lambda^{2} \\
 & \rightarrow \left| {\det D\varphi_{ij}(z)} \right| ≕ d_{ij}
\end{aligned}$$

Let $z_{ij} = Z_{ij}^{- 1}(z)$ and let
$J = \left\{ (i,j):z_{i,j} \in A \right\}$, then

$$\begin{array}{r}
P\left( A~|~\left\{ \mathcal{Z} \in B_{\delta}(z) \right\} \right) = \frac{\sum_{i,j \in \left\{ 1,2 \right\}}P\left( A \cap C_{ij\delta} \right)}{\sum_{i,j \in \left\{ 1,2 \right\}}P\left( C_{ij\delta} \right)} \\
\overset{\delta\text{ small}}{=}\frac{\sum_{(i,j) \in J}P\left( C_{ij\delta} \right)}{\sum_{i,j \in \left\{ 1,2 \right\}}P\left( C_{ij\delta} \right)} \\
 = \frac{\sum_{(i,j) \in J}\frac{1}{P\left( C_{11\delta} \right)}P\left( C_{ij\delta} \right)}{\sum_{i,j \in \left\{ 1,2 \right\}}\frac{1}{P\left( C_{11\delta} \right)}P\left( C_{ij\delta} \right)} \\
 \rightarrow \frac{\sum_{(i,j) \in J}d_{ij}}{\sum_{i,j \in \left\{ 0,1 \right\}}d_{ij}}
\end{array}$$

Now clearly,
$\left( P\left( A_{1} \times F_{2}~|~\mathcal{Z} \right)P\left( F_{1} \times A_{2}~|~\mathcal{Z} \right) \right)(z) = P\left( A~|~\mathcal{Z} \right)(z)$
if and only if $d_{12}d_{21} = d_{22}$

Since this has to hold for almost all $z$ and the determinants are
continuous, it has to hold for all $z$, we have to have

$\left| {\det D\varphi_{22}} \right| = \left| {\det D\varphi_{12}\det D\varphi_{21}} \right| = \left| {\det D\varphi_{12}D\varphi_{21}} \right|$

Let $D\varphi_{x} = \begin{pmatrix}
p_{1} & p_{2}
\end{pmatrix}$ and $D\varphi_{y} = \begin{pmatrix}
q_{1} & q_{2}
\end{pmatrix}$. Now

$$\begin{aligned}
\left| \det D\varphi_{12} \right| & = \left| \det\begin{pmatrix}
p_{1} & p_{2} \\
0 & 1
\end{pmatrix} \right| = \left| p_{1} \right| \\
\left| \det D\varphi_{21} \right| & = \left| \det\begin{pmatrix}
1 & 0 \\
q_{1} & q_{2}
\end{pmatrix} \right| = \left| q_{2} \right| \\
\left| \det D\varphi_{22} \right| & = \left| \det\begin{pmatrix}
p_{1} & p_{2} \\
q_{1} & q_{2}
\end{pmatrix} \right| = \left| p_{1}q_{2} - p_{2}q_{1} \right| \\
\left| \det D\varphi_{12}D\varphi_{21} \right| & = \left| p_{1}q_{2} \right|
\end{aligned}$$

Now
$\left| \det D\varphi_{22} \right| = \left| \det D\varphi_{12}D\varphi_{21} \right| \Leftrightarrow \forall x:p_{2}q_{1}(x) = 0 \vee p_{2}q_{1}(x) = 2p_{1}q_{2}(x)$

It is clear that you can choose $\varphi_{x}$ and $\varphi_{y}$ without
this property.

Let for example

$$\begin{array}{r}
\varphi_{x}(a,b) = ab + (1 - b)a^{2} \\
\varphi_{y}(a,b) = ab + (1 - a)b^{2}
\end{array}$$

We have $$\begin{array}{r}
D\varphi_{x} = \left( 2a(1 - b) + b,\quad a - a^{2} \right) = \begin{pmatrix}
p_{1} & p_{2}
\end{pmatrix} \\
D\varphi_{y} = \left( 2b(1 - a) + a,\quad b - b^{2} \right) = \begin{pmatrix}
q_{1} & q_{2}
\end{pmatrix}
\end{array}$$ let $x = \left( \frac{1}{2},\frac{1}{2} \right)$. Clearly,
$p_{2}q_{1} = 2ab(1 - b)\hat{}2 + b^{2}(1 - b)$ and
$p_{2}q_{1}(x) = \frac{1}{4} \neq 0$

Furthermore, $2p_{1}q_{2} = 4ba(1 - a)\hat{}2 + 2a^{2}(1 - a)$ and
$2p_{1}q_{2}(x) = \frac{1}{2} \neq \frac{1}{4}$

With these chosen, we have that $\pi_{1}{
 ⫫ {}}_{P}\pi_{2}~|~\mathcal{Z}$.

# Example 2

We will now argue, that the history
$\mathcal{H}(\pi_{1}|\mathcal{Z})$ can't exist. Let
$z_{0} \in \lbrack 0,1\rbrack^{2}$. Let $J_{0}(z) = \begin{cases}
\left\{ 1 \right\} & \text{if }z = z_{0} \\
I & \text{else }
\end{cases}$ and $J = J_{0} \circ \mathcal{Z}$. We will show
$J \vdash \pi_{1}~|~\mathcal{Z}$. Then, since the history would be a
subset of all such $J$, we would have
$\mathcal{H}(\pi_{1}~|~\mathcal{Z}) = \left\{ 1 \right\}$. But we have
just seen that $\pi_{1}{
 ⫫ {}}_{P}\pi_{2}~|~\mathcal{Z}$ and therefore the history is not a
generating index function.

Now, $$\pi_{J}(x,y) = \begin{cases}
x & \text{if }\mathcal{Z}(x,y) = z_{0} \\
(x,y) & \text{else}
\end{cases}$$ and $$\pi_{\overset{¯}{J}}(x,y) = \begin{cases}
y & \text{if }\mathcal{Z}(x,y) = z_{0} \\
() & \text{else}
\end{cases}$$

If $\left\{ z_{0} \right\}$ is a $P_{\mathcal{Z}}$ nullset, clearly
$\pi_{J}{{
 ⫫ {}}_{P}}_{\mathcal{Z}}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$, since
$\pi_{\overset{¯}{J}}$ is constant a.s. Otherwise, let
$\left\{ x_{1},x_{2} \right\} \times \left\{ y_{1},y_{2} \right\} = \mathcal{Z}^{- 1}\left( z_{0} \right)$
and ${p_{x}}_{1} = P_{1}\left( x_{1} \right)$ etc.

Now since $Z^{- 1}\left( z_{0} \right)$ is a (finite) rectangle with
positive measure, we clearly have for
$A \subseteq \left\{ x_{1},x_{2} \right\},B \subseteq \left\{ y_{1},y_{2} \right\}$
that
$P\left( A \times F_{2}|Z = z \right)P\left( F_{1} \times B|Z = z \right) = P\left( A \times B~|~Z = z \right)$
for all $z \in \mathcal{Z}(F)$. For
$A \subseteq F \smallsetminus \left\{ x_{1},x_{2} \right\} \times \left\{ y_{1},y_{2} \right\}$,
we clearly have $A{
 ⫫ {}}_{P}\varnothing~|~\mathcal{Z}$.

Since these sets $A$ are closed under intersections and generate
$\pi_{J}$ and the sets $B$ generate $\pi_{\overset{¯}{J}}$, we have
that $\pi_{J}{
 ⫫ {}}_{P}\pi_{\overset{¯}{J}}~|~\mathcal{Z}$.