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Let $(\Omega, \mathcal{F}, P)$ be a probability triplet. Let $\mathcal{G}$ and $\mathcal{H}$ be two sub-sigma-algebras of $\mathcal{F}$. Let $X:\Omega\rightarrow\mathbb{R}$ be a random variable such that $\sigma(X) \subseteq \sigma(\mathcal{G} \cup \mathcal{H})$.

Can we say there is a Borel-measurable function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and two random variables $G$ and $H$ such that $X = f(G,H)$, and $\sigma(G) \subseteq \mathcal{G}$, $\sigma(H) \subseteq \mathcal{H}$? Are there references where I can cite such a result?

If it makes the problem easier, I am also interested in the case when $\mathcal{G}$ and $\mathcal{H}$ are independent and/or when one of the sub-sigma-algebras, say $\mathcal{G}$, is generated by some random variable $R$.


Note 1: I can show it is true if there are random variables $R, S$ such that $\mathcal{G}=\sigma(R)$ and $\mathcal{H}=\sigma(S)$.

Note 2: There may be some hope by applying properties of countably generated sigma algebras, since $\sigma(X)$ is countably generated. If we define the events $B_x = \{X \leq x\}$ for each rational number $x$, I wonder if there is a way of writing a particular event $B_x$ in terms of events in $\mathcal{G}$ and $\mathcal{H}$.


Application: I have a random variable $Z$ and I want to write for some $f$: $$ E[Z|\sigma(\mathcal{G} \cup \mathcal{H})] \overset{?}{=} f(G,H)$$ This is similar in spirit to the known fact $E[Z|\sigma(Y)]=f(Y)$ for some $f$.

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  • 2
    $\begingroup$ Re your 2nd note: A $\sigma$-algebra $\mathcal{H}$ is countably generated iff $\mathcal{H} = \sigma(f)$ for some mapping $f$. Combining this with your first remark, this should give the result for countably generated $\sigma$-algebras. I wouldn't expect the result to hold for $\sigma$-algebras which are nout countably generated but it's just a feeling $\endgroup$
    – saz
    Commented Jul 12, 2020 at 6:46
  • $\begingroup$ @saz : Thanks for your thoughts. FYI I've edited the problem to show my application. $\endgroup$
    – Michael
    Commented Jul 13, 2020 at 15:36
  • $\begingroup$ At a glance, I would say that the result cannot be proved unless the sigma-algebras $\mathcal{G}$ and $\mathcal{H}$ are generated by one random variable (namely your $G$ and $H$). If this condition holds, then it is easy, because in that case you can apply Doob’s measurability theorem to the random vector $(G,H)$ and have your result done. But in general, I suspect that the result is simply false. $\endgroup$
    – Logos
    Commented Jul 13, 2020 at 16:05
  • $\begingroup$ @saz : Actually I believe I have proved my conjecture is true; see my (second) answer below. Any comments are welcome. $\endgroup$
    – Michael
    Commented Jul 20, 2020 at 9:54
  • $\begingroup$ @Logos : As I mention in my above comment to saz, any comments you might have on my answer below are welcome. $\endgroup$
    – Michael
    Commented Jul 20, 2020 at 9:54

3 Answers 3

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The desired structure is true!

Setup

We are given probability triplet $(\Omega, \mathcal{F}, P)$. Suppose $\mathcal{G}$ and $\mathcal{H}$ are two sub-sigma-algebras of $\mathcal{F}$. Suppose $X:\Omega\rightarrow\mathbb{R}$ is $\sigma(\mathcal{G} \cup \mathcal{H})$-measurable, meaning that for every Borel set $B \subseteq \mathbb{R}$ we have $\{\omega \in \Omega : X(\omega) \in B\} \in \sigma(\mathcal{G} \cup \mathcal{H})$.

Proposition 1

There exists a Borel-measurable function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and two random variables $G:\Omega\rightarrow[0,1)$ and $H:\Omega\rightarrow[0,1)$, where $G$ is $\mathcal{G}$-measurable and $H$ is $\mathcal{H}$-measurable, such that $X=f(G,H)$.

The proof of Prop 1 is developed below using (i) a proposition about Borel-measurability of manipulations of decimal expansions; (ii) a multiplicative class theorem.

Notation

If $\mathbb{S}$ is a set of functions $Y:\Omega\rightarrow\mathbb{R}$ then $$ \sigma(\mathbb{S}) = \sigma(\cup_{Y \in \mathbb{S}} \sigma(Y))$$ where $\sigma(Y)$ is the sigma-algebra generated by $Y$.

Proposition 2

Let $x \in [0,1)$ and let $x=0.x_1x_2x_3...$ be the unique decimal expansion, where $x_i \in \{0, 1, ..., 9\}$ and there is no infinite tail of 9s. For $x \in [0,1)$ let $\phi_n(x)=x_n$ denote the $n$th digit of the unique expansion of $x$. Then $\phi_n(x)$ is Borel-measurable. Let $\{n_i\}_{i=1}^{\infty}$ be an infinite sequence of increasing positive integers. Then the map $h:[0,1)\rightarrow \mathbb{R}$ defined by $$ h(x) = \sum_{i=1}^{\infty} \phi_{n_i}(x)10^{-i}$$ is Borel-measurable.

For example if $\{n_i\} = \{2, 4, 6, 8, ...\}$ then $$ x = 0.14159265358979323 \implies h(x) = 0.45255992...$$

Proof of Prop 2:

It is not difficult to see that $\phi_n(x)$ is Borel-measurable for all positive integers $n$. Then $\sum_{i=1}^k \phi_{n_i}(x)10^{-i}$ is Borel-measurable for each positive integer $k$. The limit of these Borel-measurable functions is also Borel-measurable. $\Box$

Multiplicative class theorem

We use Theorem 11.2 from Analysis Tools with Examples by B. K. Driver: http://www.math.ucsd.edu/~bdriver/DRIVER/Book/anal.pdf

Theorem 11.2 : Let $\Omega$ be a nonempty set. Let $\mathbb{V}$ be set of bounded functions $Y:\Omega\rightarrow\mathbb{R}$. Assume

  1. $\mathbb{V}$ contains all constant functions.

  2. $\mathbb{V}$ is a vector subspace: If $Y, Z \in \mathbb{V}$ and $a,b\in \mathbb{R}$ then $aY+bZ \in \mathbb{V}$.

  3. $\mathbb{V}$ is closed with respect to bounded convergence. This means that if $M$ is a finite real number and $\{Y_n\}_{n=1}^{\infty}$ is a sequence of functions in $\mathbb{V}$ that satisfy

  • $|Y_n(\omega)|\leq M$ for all $\omega \in \Omega$ and $n \in \{1, 2 ,3,...\}$.

  • $\lim_{n\rightarrow\infty} Y_n(\omega) = Y(\omega)$ for all $\omega \in \Omega$, for some function $Y:\Omega\rightarrow\mathbb{R}$.

then $Y \in \mathbb{V}$.

  1. There is some "multiplicative system" $\mathbb{M}$ such that $\mathbb{M} \subseteq \mathbb{V}$. Specifically, $\mathbb{M}$ is a multiplicative system if $Y \in \mathbb{M}$ and $Z \in \mathbb{M}$ imply $YZ \in \mathbb{M}$.

Then $\mathbb{V}$ contains all bounded and $\sigma(\mathbb{M})$-measurable functions. Note that $\sigma(\mathbb{M}) = \sigma(\cup_{Y \in \mathbb{M}}\sigma(Y))$.

Proof of Prop 1

It suffices to prove the result for the case $X$ is bounded, since unbounded $X$ can be mapped to bounded $Y$ through an invertible continuous function (as in an early comment by John Dawkins).

Define $\mathbb{V}$ as the set of all bounded functions $Y:\Omega\rightarrow\mathbb{R}$ such that $Y=f(G,H)$ for some Borel-measurable function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and some random variables $G:\Omega\rightarrow[0,1)$ and $H:\Omega\rightarrow[0,1)$ such that $G$ is $\mathcal{G}$-measurable and $H$ is $\mathcal{H}$-measurable.

We first show that $\mathbb{V}$ satisfies the properties needed for Theorem 11.2:

  1. Indeed $\mathbb{V}$ contains all constant functions.

  2. Indeed $\mathbb{V}$ is a vector subspace: Let $Y_1$ and $Y_2$ be functions in $\mathbb{V}$ and let $a_1, a_2$ be scalars. We want to show $a_1 Y_1 + a_2 Y_2 \in \mathbb{V}$. We know $Y_1 = f_1(G_1,H_1)$ and $Y_2 = f_2(G_2,H_2)$ for random variables $G_1$, $G_2$ being $\mathcal{G}$-measurable and taking values in $[0,1)$; random variables $H_1, H_2$ being $\mathcal{H}$-measurable and taking values in $[0,1)$; $f_1, f_2$ Borel-measurable. For each $\omega \in \Omega$ write $G_1(\omega)$ and $G_2(\omega)$ in their unique decimal expansions: \begin{align} G_1(\omega) &= \sum_{i=1}^{\infty} A_i(\omega)10^{-i} \\ G_2(\omega) &= \sum_{i=1}^{\infty} B_i(\omega)10^{-i} \end{align} where $A_i(\omega), B_i(\omega) \in \{0, 1, ..., 9\}$ and the sequences $\{A_i(\omega)\}_{i=1}^{\infty}$ and $\{B_i(\omega)\}_{i=1}^{\infty}$ do not have an infinite tail of 9s. Observe that $A_i(\omega)$ and $B_i(\omega)$ are $\mathcal{G}$-measurable. Define $G(\omega) \in [0,1)$ by interlacing the digits of the expansion: $$ G(\omega)=0.A_1(\omega)B_1(\omega)A_2(\omega)B_2(\omega)A_3(\omega)B_3(\omega)...$$ We can obtain $G$ from $(G_1,G_2)$ and vice-versa: $$ G \leftrightarrow (G_1,G_2)$$ Observe that $G$ is $\mathcal{G}$-measurable because for any sequence $\{r_i\}_{i=1}^{\infty}$ with $r_i \in \{0, 1, ..., 9\}$ we have $$ G \leq 0.r_1r_2r_3r_4...$$ if and only if $$ \{A_1<r_1\} \cup (\{A_1=r_1\} \cap \{B_1<r_2\}) \cup (\{A_1=r_1\}\cap\{B_1=r_2\}\cap\{A_2<r_3\}) \cup ...$$ and the right-hand-side is a countable union of events in $\mathcal{G}$. Similarly we can construct $H(\omega) \in [0,1)$ that is $\mathcal{H}$-measurable such that $$ H \leftrightarrow (H_1,H_2)$$ Then \begin{align} a_1Y_1 + a_2Y_2 &= a_1f_1(G_1,H_1) + a_2f_2(G_2,H_2) \\ &= a_1f_1(h_1(G), h_1(H)) + a_2f_2(h_2(G), h_2(H)) \end{align} where $h_1(G)=G_1$, $h_1(H)=H_1$; $h_2(G)=G_2$, $h_2(H)=H_2$, where $h_1$ is the function that takes $G$ and extracts $G_1$ (which is Borel-measurable by Prop 2) and $h_2$ the function that takes $G$ and extracts $G_2$.

  3. Suppose $\{Y_n\}_{n=1}^{\infty}$ are functions in $\mathbb{V}$ such that $|Y_n(\omega)|\leq M$ for all $n$ and all $\omega \in \Omega$ (for some $M$ such that $0<M<\infty)$ and $Y_n\rightarrow Y$ for some $Y:\Omega\rightarrow\mathbb{R}$. We want to show $Y \in \mathbb{V}$. Clearly $|Y(\omega)|\leq M$ for all $\omega\in \Omega$. Also, we have $Y_n=f_n(G_n,H_n)$ where $G_n$ is $\mathcal{G}$-measurable and $H_n$ is $\mathcal{H}$-measurable, both $G_n$ and $H_n$ taking values in $[0,1)$. Form $G$ by interlacing the digits of the decimal expansions of $\{G_n\}_{n=1}^{\infty}$. Thus $$ G \leftrightarrow (G_1, G_2, G_3, ...)$$ and $G$ is $\mathcal{G}$-measurable. Similarly we can form $H$ that is $\mathcal{H}$-measurable and $$ H \leftrightarrow (H_1, H_2, H_3, ...)$$ Then for all $\omega$ we have $$Y(\omega) = \lim_{n\rightarrow\infty} f_n(G_n(\omega), H_n(\omega)) = \lim_{n\rightarrow\infty} f_n(\psi_n(G), \psi_n(H))$$ where $\psi_n(G)=G_n$ and $\psi_n(H)=H_n$, and $\psi_n(\cdot)$ is the Borel-measurable function that takes $G$ and extracts $G_n$. Define $f:\mathbb{R}^2\rightarrow\mathbb{R}$ by $$ f(x,y) = \limsup_{n\rightarrow\infty}[ f_n(\psi_n(x),\psi_n(y))]_{-2M}^{2M}$$ where $[z]_{-2M}^{2M}$ projects the real number $z$ onto the interval $[-2M, 2M]$. The limsup of bounded and Borel-measurable functions is bounded and Borel-measurable, and so $f$ is a real-valued Borel-measurable function and $Y=f(G,H)$.

  4. Define $\mathbb{M}$ as the set of all bounded functions $Y:\Omega\rightarrow\mathbb{R}$ such that $Y=g(G)h(H)$ for some Borel-measurable functions $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow\mathbb{R}$, where $G$ is $\mathcal{G}$-measurable, $H$ is $\mathcal{H}$-measurable, and both $G$ and $H$ take values in $[0,1)$. By similar arguments it holds that $\mathbb{M}$ is a multiplicative system. Indeed, take $Y_1=g_1(G_1)h_1(H_1), Y_2=g_2(G_2)h_2(H_2)$ in $\mathbb{M}$, then $$Y_1Y_2 = g_1(G_1)g_2(G_2)h_1(H_1)h_2(H_2) = g_1(\phi_1(G))g_2(\phi_2(G))h_1(\phi_1(H))h_2(\phi_2(H))$$ where $\phi_1(G) = G_1$ and $\phi_2(G)=G_2$.

By Theorem 11.2 we know that $\mathbb{V}$ contains all bounded $\sigma(\mathbb{M})$-measurable functions.

It remains to show that $\sigma(\mathcal{G}\cup\mathcal{H}) \subseteq \sigma(\mathbb{M})$. This would imply that our bounded random variable $X$ is $\sigma(\mathbb{M})$-measurable and so $X \in \mathbb{V}$, meaning that $X=f(G,H)$ for some $G:\Omega\rightarrow\mathbb{R}$ and $H:\Omega\rightarrow\mathbb{R}$, both taking values in $[0,1)$, such that $G$ is $\mathcal{G}$-measurable and $H$ is $\mathcal{H}$-measurable.

Take any set $A \in \mathcal{G}$. Define $G=(1/2)1_A$ and $H=(1/2)$. Then $X=GH = (1/4)1_A$ is in $\mathbb{M}$. So $\sigma(X) \subseteq \sigma(\mathbb{M})$ and so $A \in \sigma(\mathbb{M})$. Similarly if $B \in \mathcal{H}$ then $B \in \sigma(\mathbb{M})$. Thus $\mathcal{G} \subseteq \sigma(\mathbb{M})$ and $\mathcal{H} \subseteq \sigma(\mathbb{M})$ and so $\sigma(\mathcal{G} \cup \mathcal{H}) \subseteq \sigma(\mathbb{M})$.

$\Box$

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  • $\begingroup$ It follows by induction that if $\mathcal{G}_i$ are sub-sigma-algebras of $\mathcal{F}$ for $i \in \{1, ..., n\}$ and $X:\Omega\rightarrow\mathbb{R}$ is $\sigma(\cup_{i=1}^n \mathcal{G}_i)$-measurable then $X=f(G_1, ..., G_n)$ for some Borel-measurable function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ and some random variables $G_i:\Omega\rightarrow\mathbb{R}$ that are $\mathcal{G}_i$-measurable for each $i \in \{1, ..., n\}$. $\endgroup$
    – Michael
    Commented Jul 20, 2020 at 11:40
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What is true, and this may be sufficient for your purposes, is that under the stated conditions there is a $\mathcal G\otimes\mathcal H$-measurable map $Z:\Omega\times\Omega\to\Bbb R$ such that $X(\omega) = Z(\omega,\omega)$ for all $\omega\in\Omega$. (And conversely, because $\omega\mapsto(\omega,\omega)$ is $\sigma(\mathcal G\cup\mathcal H)/\mathcal G\otimes\mathcal H$-measurable.)

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  • $\begingroup$ I tried proving your claim by constructing the (overly simple?) map $Z(\omega_1, \omega_2) = X(\omega_1)$ but it is not clear how to show $\{(\omega_1,\omega_2): X(\omega_1) \leq x\} \in \mathcal{G} \otimes \mathcal{H}$. It would be true if $(\omega, z) \rightarrow \omega$ is $\mathcal{G}\otimes \mathcal{H} / \sigma(\mathcal{G}\cup \mathcal{H})$ measurable but this does not seem obvious. Embarassingly I also tried showing $\omega \rightarrow (\omega, \omega)$ is $\sigma(\mathcal{G}\cup \mathcal{H})/\mathcal{G}\otimes\mathcal{H}$ measurable but didn't get anywhere for 10 minutes of effort. $\endgroup$
    – Michael
    Commented Jul 15, 2020 at 17:15
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    $\begingroup$ I had in mind a monotone class argument. Assume $X$ is bounded. Let $\Bbb V$ be the vector space of bdd $\sigma(\mathcal G\cup\mathcal H)$-meas. maps from $\Omega$ to $\Bbb R$ with the asserted property. Then $\Bbb H$ is closed under bdd monotone convergence and contains the (multiplicative) class $\Bbb K$ of all non-neg. maps of the form $\omega\mapsto G(\omega)H(\omega)$, where $G$ is $\mathcal G$-meas. and $H$ is $\mathcal H$-meas., both bdd. (Take $Z(\omega,\omega')=G\omega)H(\omega')$.) As such, $\Bbb V$ contains all bounded $\sigma(\mathcal G\cup\mathcal H)$-meas. maps, including $X$. $\endgroup$ Commented Jul 15, 2020 at 17:53
  • $\begingroup$ I think $\mathbb{V}$ changed to $\mathbb{H}$ (typo)? So I think $\mathbb{V}$ is the vector space of bounded $\sigma(\mathcal{G}\cup\mathcal{H})$-meas maps $Y:\Omega\rightarrow\mathbb{R}$ such that $Y(\omega)=Z(\omega,\omega)$ for some $\mathcal{G}\otimes \mathcal{H}$-measurable map $Z:\Omega^2\rightarrow\mathbb{R}$, and I agree $\mathbb{V}$ contains all maps of the form $G(\omega)H(\omega)$ with bounded $G$ being $\mathcal{G}$-meas and bounded $H$ being $\mathcal{H}$-meas (regardless of nonneg). $\endgroup$
    – Michael
    Commented Jul 15, 2020 at 20:26
  • $\begingroup$ (for me to check later): The multiplicative class statement may have something to do with Theorem 31 in these notes I found: fmf.uni-lj.si/~vidmarm/Dynkin_and_pi_systems.pdf $\endgroup$
    – Michael
    Commented Jul 15, 2020 at 20:38
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    $\begingroup$ It is easy enough to reduce to the case of bounded $X$ by considering $\arctan(X)$, for example. $\endgroup$ Commented Jul 17, 2020 at 15:07
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This answer fills in details of the John Dawkins answer (which gave no proof) and comments (which gave a proof sketch). I ended up changing things to extended-real-valued functions to ensure the proof steps work.

Setup

We have a probability triplet $(\Omega, \mathcal{F}, \mathbb{R})$. We have $\mathcal{G}, \mathcal{H}$ that are sub-sigma-algebras of $\mathcal{F}$, we have a random variable $X:\Omega\rightarrow\mathbb{R}$ such that $\sigma(X) \subseteq \sigma(\mathcal{G} \cup \mathcal{H})$.

Define the extended real numbers: $$ \mathbb{E} = \mathbb{R} \cup \{\infty\} \cup \{-\infty\}$$

Claim

There is a $\sigma(\mathcal{G}\otimes \mathcal{H})$-measurable function $Z:\Omega^2\rightarrow \mathbb{E}$ such that $X(\omega)=Z(\omega,\omega)$ for all $\omega \in \Omega$.

Proof:

We will prove for the special case of $X$ bounded. The boundedness assumption can be removed as described by comments to John's answer.


I use this link to "Analysis Tools with Examples" by B. K. Driver:

http://www.math.ucsd.edu/~bdriver/DRIVER/Book/anal.pdf

Theorem 11.2 states this:

Let $\Omega$ be a nonempty set. Let $\mathbb{H}$ be a vector subspace of bounded functions $X:\Omega\rightarrow\mathbb{R}$. Assume

  1. $\mathbb{H}$ contains all constant functions.

  2. $\mathbb{H}$ is a vector subspace: If $X, Y \in \mathbb{H}$ and $a,b\in \mathbb{R}$ then $aX+bY \in \mathbb{H}$.

  3. If $\{X_n\}_{n=1}^{\infty}$ is a sequence of functions in $\mathbb{H}$ such that there is an $M<\infty$ such that:

  • $|X_n(\omega)|\leq M$ for all $\omega \in \Omega$ and $n \in \{1, 2 ,3,...\}$.

  • $\lim_{n\rightarrow\infty} X_n(\omega) = X(\omega)$ for all $\omega \in \Omega$, for some function $X:\Omega\rightarrow\mathbb{R}$.

Then $X \in \mathbb{H}$.

  1. There is some "multiplicative system" $\mathbb{M}$ such that $\mathbb{M} \subseteq \mathbb{H}$. Specifically, $\mathbb{M}$ is a multiplicative system if $X \in \mathbb{M}$ and $Y \in \mathbb{M}$ imply $XY \in \mathbb{M}$.

Then $\mathbb{H}$ contains all bounded and $\sigma(\mathbb{M})$-measurable functions.


Notational note: If $\mathbb{M}$ is a set of functions $X:\Omega\rightarrow\mathbb{R}$ then $\sigma(\mathbb{M})$ is defined $$ \sigma(\mathbb{M}) = \sigma(\cup_{X \in \mathbb{M}} \sigma(X))$$ where $\sigma(X)$ is the sigma algebra generated by $X:\Omega\rightarrow\mathbb{R}$.


Following John's suggestions (modifying them to extended real-valued functions): Define $\mathbb{H}$ as the set of all bounded functions $Y:\Omega\rightarrow\mathbb{R}$ such that $\sigma(Y) \subseteq \sigma(\mathcal{G} \cup \mathcal{H})$ and $Y(\omega) = Z(\omega,\omega)$ for all $\omega \in \Omega$, where $Z:\Omega^2 \rightarrow\mathbb{E}$ is some function that satisfies $\sigma(Z) \subseteq \mathcal{F} \otimes \mathcal{G}$. Let us first show this satisfies the requirements of Theorem 11.2:

  1. Indeed $\mathbb{H}$ contains all constant functions: Fix $c \in \mathbb{R}$. If $Y:\Omega\rightarrow\mathbb{R}$ is defined by $Y(\omega) = c$ for all $\omega \in \Omega$ then $\sigma(Y) = \{\Omega, \phi\} \subseteq \sigma(\mathcal{G} \cup \mathcal{H})$. And we can write $Y(\omega) = Z(\omega, \omega)$ for $Z:\Omega^2\rightarrow\mathbb{R}$ defined by $Z(\omega, v) = c$ for all $(\omega, v) \in \Omega^2$, and indeed $\sigma(Z) \subseteq \mathcal{F} \otimes \mathcal{G}$.

  2. Indeed $\mathbb{H}$ is a vector space.

  3. Suppose $M$ is a constant and $\{X_n\}_{n=1}^{\infty}$ is a sequence of functions in $\mathbb{H}$ that satisfy $|X_n(\omega)|\leq M$ for all $n \in \{1, 2, 3, ...\}$ and all $\omega \in \Omega$. Suppose there is a function $X:\Omega\rightarrow\mathbb{R}$ such that $X_n(\omega)\rightarrow X(\omega)$ for all $\omega \in \Omega$. We want to show $X \in \mathbb{H}$. Since all $X_n$ functions are uniformly bounded by $M$, the limit function $X$ must be uniformly bounded by $M$. Next, we know the pointwise limit of $\sigma(\mathcal{G} \cup \mathcal{H})$-measurable functions is also $\sigma(\mathcal{G} \cup \mathcal{H})$-measurable, so $X$ is $\sigma(\mathcal{G} \cup \mathcal{H})$-measurable. Finally, we know for each $n$ that $X_n(\omega) = Z_n(\omega, \omega)$ for some $(\mathcal{G}\otimes \mathcal{H})$-measurable functions $Z_n:\Omega^2\rightarrow\mathbb{E}$. There is a theorem that says the function $\limsup_{n\rightarrow\infty} Z_n(\omega,v)$ is an extended real-valued function that is also $(\mathcal{G}\otimes \mathcal{H})$-measurable. And of course $X(\omega) = \limsup_{n\rightarrow\infty} Z_n(\omega, \omega)$ for all $\omega \in \Omega$.

  4. Let $\mathbb{M}$ be the set of all bounded functions $X:\Omega\rightarrow\mathbb{R}$ such that $X(\omega) = G(\omega)H(\omega)$ for some $\mathcal{G}$-measurable function $G:\Omega\rightarrow\mathbb{R}$ and some $\mathcal{H}$-measurable function $H:\Omega\rightarrow\mathbb{R}$.Then $\mathbb{M}$ is a multiplicative system because the product of $\mathcal{G}$-measurable functions is $\mathcal{G}$-measurable and the product of $\mathcal{H}$-measurable functions is $\mathcal{H}$-measurable.

From Theorem 11.2 we conclude that $\mathbb{H}$ contains all bounded $\sigma(\mathbb{M})$-measurable functions.


We now require to prove that $\sigma(\mathcal{G} \cup \mathcal{H}) \subseteq \sigma(\mathbb{M})$. This would imply that our bounded random variable $X$ is $\sigma(\mathbb{M})$-measurable and so $X \in \mathbb{H}$, meaning that $X(\omega) = Z(\omega, \omega)$ for some $(\mathcal{G}\otimes\mathcal{H})$-measurable function $Z:\Omega^2\rightarrow\mathbb{E}$.

Take any set $A \in \mathcal{G}$. Define $G=1_A$ and $H=1$. Then $X=GH = 1_A$ is in $\mathbb{M}$. So $\sigma(X) \subseteq \sigma(\mathbb{M})$ and so $A \in \sigma(\mathbb{M})$. Similarly if $B \in \mathcal{H}$ then $B \in \sigma(\mathbb{M})$. Thus $\mathcal{G} \subseteq \sigma(\mathbb{M})$ and $\mathcal{H} \subseteq \sigma(\mathbb{M})$ and so $\sigma(\mathcal{G} \cup \mathcal{H}) \subseteq \sigma(\mathbb{M})$.

$\Box$

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  • $\begingroup$ Of course once it holds for $(\mathcal{G}\otimes \mathcal{H})$-measurable $Z:\Omega^2\rightarrow\mathbb{E}$ it also holds for $Z:\Omega^2\rightarrow\mathbb{R}$ by defining $$\tilde{Z}(\omega, v)= \left\{\begin{array}{cc} Z(\omega, v) & \mbox{ if $|Z|\neq \infty$} \\ 0 & \mbox{ else} \end{array}\right.$$ $\endgroup$
    – Michael
    Commented Jul 20, 2020 at 2:43

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