# If $X$ is $\sigma(\mathcal{G} \cup \mathcal{H})$-measurable can we write $X=f(G,H)$?

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 $$f$$.

• 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
– saz
Commented Jul 12, 2020 at 6:46
• @saz : Thanks for your thoughts. FYI I've edited the problem to show my application. Commented Jul 13, 2020 at 15:36
• 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. Commented Jul 13, 2020 at 16:05
• @saz : Actually I believe I have proved my conjecture is true; see my (second) answer below. Any comments are welcome. Commented Jul 20, 2020 at 9:54
• @Logos : As I mention in my above comment to saz, any comments you might have on my answer below are welcome. Commented Jul 20, 2020 at 9:54

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 $$\{A_1 and the right-hand-side is a countable union of events in $$\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 $$0 and $$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 $$\sigma(\mathcal{G} \cup \mathcal{H}) \subseteq \sigma(\mathbb{M})$$.

$$\Box$$

• 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 $i \in \{1, ..., n\}$. Commented Jul 20, 2020 at 11:40

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.)

• 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. Commented Jul 15, 2020 at 17:15
• 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 $X$. Commented Jul 15, 2020 at 17:53
• 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). Commented Jul 15, 2020 at 20:26
• (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 Commented Jul 15, 2020 at 20:38
• It is easy enough to reduce to the case of bounded $X$ by considering $\arctan(X)$, for example. Commented Jul 17, 2020 at 15:07

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 $$\sigma(\mathcal{G} \cup \mathcal{H}) \subseteq \sigma(\mathbb{M})$$.

$$\Box$$

• 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 $$\tilde{Z}(\omega, v)= \left\{\begin{array}{cc} Z(\omega, v) & \mbox{ if |Z|\neq \infty} \\ 0 & \mbox{ else} \end{array}\right.$$ Commented Jul 20, 2020 at 2:43