Theorem: Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be a integrable random variable.
Let $\map \sigma X$ be the $\sigma$-algebra generated by $X$.
Let $\HH \subseteq \Sigma$ be a sub-$\sigma$-algebra that is independent of $\map \sigma {\map \sigma X, \GG}$, the $\sigma$-algebra generated by $\map \sigma X \cup \GG$.
Let $\map \sigma {\GG, \HH}$ be the $\sigma$-algebra generated by $\GG \cup \HH$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $\expect {X \mid \map \sigma {\GG, \HH} }$ be a version of the conditional expectation of $X$ given $\map \sigma {\GG, \HH}$.
Then: 
:$\expect {X \mid \map \sigma {\GG, \HH} } = \expect {X \mid \GG}$ almost surely.

Proof:
First take $X$ to be a non-negative random variable.
Let:
:$\SS = \set {G \cap H : G \in \GG, \, H \in \HH}$
We aim to apply Uniqueness of Measures to a suitable measure with $\SS$. 
We start by showing that $\Omega \in \SS$, $\SS$ is a $\pi$-system, and that $\SS$ generates $\map \sigma {\GG, \HH}$.
First note that:
:$\Omega = \Omega \cap \Omega \in \SS$ 
We now show that $\SS$ is a $\pi$-system.
Let $G_1 \cap H_1, G_2 \cap H_2, \ldots, G_n \cap H_n \in \SS$ with $G_i \in \GG$ and $H_i \in \HH$ for each $i$.
We have from Intersection is Associative:
:$\ds \bigcap_{i \mathop = 1}^n \paren {G_i \cap H_i} = \paren {\bigcap_{i \mathop = 1}^n G_i} \cap \paren {\bigcap_{i \mathop = 1}^n H_i}$
Since $\GG$ and $\HH$ are $\sigma$-algebras we have:
:$\ds \bigcap_{i \mathop = 1}^n G_i \in \GG$
and:
:$\ds \bigcap_{i \mathop = 1}^n H_i \in \HH$ 
so:
:$\ds \bigcap_{i \mathop = 1}^n \paren {G_i \cap H_i} \in \SS$ 
We now show that $\SS$ generates $\map \sigma {\GG, \HH}$.
By the definition of the $\sigma$-algebra generated by $\GG \cup \HH$, we just need to show that: 
:$\GG \subseteq \SS$ and $\HH \subseteq \SS$
and:
:$\SS \subseteq \map \sigma {\GG, \HH}$
Note that for each $G \in \GG$ we have $G = G \cap \Omega \in \SS$, and for each $H \in \HH$ we have $H = H \cap \Omega \in \SS$, so we have:
:$\GG \subseteq \SS$ and $\HH \subseteq \SS$
Now let $G \in \GG$ and $H \in \GG$ so that $G \cap H \in \SS$. 
Then $G, H \in \map \sigma {\GG, \HH}$, so $G \cap H \in \map \sigma {\GG, \HH}$ from the definition of a $\sigma$-algebra. 
So we indeed have: 
:$\SS \subseteq \map \sigma {\GG, \HH}$
and:
:$\map \sigma {\SS} = \map \sigma {\GG, \HH}$
Now define the measure with density $\mu_1$ by: 
:$\ds \map {\mu_1} A = \int_A X \rd \Pr$
for $A \in \map \sigma {\GG, \HH}$.
Also define: 
:$\ds \map {\mu_2} A = \int_A \expect {X \mid \GG} \rd \Pr$
Since $X$ and $\expect {X \mid \GG}$ are integrable, these are finite measures.
Therefore, by Uniqueness of Measures, if we can show: 
:$\map {\mu_1} A = \map {\mu_2} A$ for all $A \in \SS$
we will have that $\mu_1 = \mu_2$. 
This will show that $\expect {X \mid \GG}$ is a version of the conditional expectation of $X$ given $\map \sigma {\GG, \HH}$.
We will then have, by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra:
:$\expect {X \mid \map \sigma {\GG, \HH} } = \expect {X \mid \GG}$ almost surely.
Let $G \in \GG$ and $H \in \HH$ so that $G \cap H \in \SS$. 
Then we have, by Characteristic Function of Intersection:
:$\map {\mu_1} {G \cap H} = \expect {X \cdot \chi_{G \cap H} } = \expect {X \cdot \chi_G \chi_H}$
From Characteristic Function Measurable iff Set Measurable, we have: 
:$\chi_G$ is $\GG$-measurable
and so:
:$\chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.
From Pointwise Product of Measurable Functions is Measurable, we have: 
:$X \cdot \chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.
Now from Expectation of Product of Independent Random Variables is Product of Expectations, we have: 
:$\expect {X \cdot \chi_G \chi_H} = \expect {X \cdot \chi_G} \expect {\chi_H}$
So, from Integral of Characteristic Function:
:$\map {\mu_1} {G \cap H} = \expect {X \cdot \chi_G} \map \Pr H$
As to $\mu_2$, we have from Characteristic Function of Intersection: 
:$\map {\mu_2} {G \cap H} = \expect {\expect {X \mid \GG} \cdot \chi_{G \cap H} } = \expect {\expect {X \mid \GG} \cdot \chi_G \chi_H}$
Since: 
:$\expect {X \mid \GG}$ is $\GG$-measurable
and again:
:$\chi_G$ is $\GG$-measurable
so by Pointwise Product of Measurable Functions is Measurable:
:$\expect {X \mid \GG} \cdot \chi_G$ is $\GG$-measurable
and so:
:$\expect {X \mid \GG} \cdot \chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.
Now, from Expectation of Product of Independent Random Variables is Product of Expectations we have:
:$\expect {\expect {X \mid \GG} \cdot \chi_G \chi_H} = \expect {\expect {X \mid \GG} \cdot \chi_G} \expect {\chi_H}$
Then from Integral of Characteristic Function we have:
:$\expect {\expect {X \mid \GG} \cdot \chi_G} \expect {\chi_H} = \expect {\expect {X \mid \GG} \cdot \chi_G} \map \Pr H$ 
From the definition of a version of the conditional expectation of $X$ given $\GG$, we have:
:$\expect {\expect {X \mid \GG} \cdot \chi_G} = \expect {X \cdot \chi_G} \map \Pr H = \map {\mu_1} A$ 
So we have shown that: 
:$\map {\mu_1} A = \map {\mu_2} A$ for each $A \in \SS$
Hence the result.
{{qed}}