Why $\mathbb{E}[X\mid \sigma(\mathcal{H},\mathcal{E})]=\mathbb{E}[X\mid \mathcal{H}]$? My problem:
Suppose $\mathcal{E}$ and $\mathcal{H}$ are sub-$\sigma$-algebras of the $\sigma$-algebra $\mathcal{F}$.
Let $X \in L^1(\Omega,\mathcal{F},\mathbb{P})$ and $\sigma(X)=\{X^{-1}(A): A \in \mathcal{B}(\mathbb{R})   \}$. Suppose that $\mathcal{E}$ is independent from $\sigma(\mathcal{H},\sigma(X))$.
Then $$\mathbb{E}[X\mid \sigma(\mathcal{H},\mathcal{E})]=\mathbb{E}[X\mid \mathcal{H}]$$
My attempt:
I tried using the characterisation $\mathbb{E}[XZ]=\mathbb{E}[\mathbb{E}[X\mid \mathcal{H}]Z]$ for all $\mathcal{H}$-measurable and bounded random variable or $\mathbb{E}[XZ]=\mathbb{E}[\mathbb{E}[X\mid \sigma(\mathcal{H},\sigma(X))]Z]$ for all $\sigma(\mathcal{H},\sigma(X))$-measurable and bounded random variable.
 A: This is a ell known result by Doob.
Theorem: Let $\mathscr{A}$, $\mathscr{B}$ and $\mathscr{C}$ be sub--$\sigma$--algebras of $\mathscr{F}$.  $\mathscr{A}\perp_\mathscr{C} \mathscr{B}$ iff
$$
\begin{align}
\Pr[A|\sigma(\mathscr{C},\mathscr{B})]=\Pr[A|\mathscr{C}]\tag{1}\label{doob-independence}
\end{align}
$$
for all $A\in \mathscr{A}$.
Here is a shot proof:
Suppose that $\mathscr{A}$ and $\mathscr{B}$ are conditional independent given $\mathscr{C}$, that is
$$
\Pr[A\cap B|\mathscr{C}]=\Pr[A|\mathscr{C}] \Pr[B|\mathscr{C}]
$$
for all $A\in \mathscr{A}$ and $B\in \mathscr{B}$. Then, for any $A\in\mathscr{A}$, $\mathscr{B}$ and $C\in\mathscr{C}$
we have
$$
\begin{align}
\Pr\big[A\cap\big(C\cap B)\big]&=\Pr\big[ \mathbb{1}_C\Pr[A\cap B|\mathscr{C}]\big]=
 \Pr\big[\mathbb{1}_C\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]\big]\\
&= \Pr\big[\Pr[A|\mathscr{C}]\Pr[B\cap C|\mathscr{C}]\big]=
\Pr\Big[\Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_{B\cap C}\big|\mathscr{C}\big]\Big]\\
&= \Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_{B\cap C}\big]
\end{align}
$$
Since $\sigma(\mathscr{B},\mathscr{C})=\sigma\Big(\{B\cap C: B\in\mathscr{B}, C\in\mathscr{C}\}\Big)$, a monotone class argument shows that
$$
\begin{align}
\Pr[A\cap H]=\Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_H \big]
\end{align}
$$
for all $H\in\sigma(\mathscr{B},\mathscr{C})$. This means that
$$
\Pr[A|\sigma(\mathscr{B},\mathscr{C})]=\Pr[A|\mathscr{C}]
$$
Conversely, suppose that $\eqref{doob-independence}$ holds. For any $A\in\mathscr{A}$ and $B\in\mathscr{B}$ we have
\begin{align*}
\Pr[A\cap B|\mathscr{C}]=\Pr\Big[\mathbb{1}_{B}\Pr[A|\sigma(\mathscr{B},\mathscr{C})]\Big| \mathscr{C}\Big]=
\Pr\Big[\mathbb{1}_B\Pr[A|\mathscr{C}]\Big|\mathscr{C}\Big]
=\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]
\end{align*}
This shows that $\mathscr{A}$ and $\mathscr{B}$ are independent given $\mathscr{C}$.
The extension to random variables is done by expanding first to simple functions and then by the usual monotone approximation by simple functions.
