Is $\mathbb E [\mathbb{1}_A \mid \mathcal{D} \vee \mathcal{G}]$ $\mathcal{G}$-measurable? My lecture notes define conditional expectation and independence as follows:



*

*Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathcal{G}$ be a sub-sigma field of $\mathcal{F}$. If $X$ is an integrable random variable, then the conditional expectation of $X$ given $\mathcal{G}$ is any integrable random variable $Z$ which satisfies the following two properties:


(CE1) $Z$ is $\mathcal{G}$-measurable.
(CE2) $$\forall \Lambda \in \mathcal{G}: \int_{\Lambda} Z \, d \mathbb{P}=\int_{\Lambda} X \, d \mathbb{P}$$
We denote $Z$ by $\mathbb{E}[X | \mathcal{G}]$.


*A finite family $\mathcal{G}_{1}, \ldots, \mathcal{G}_{n}$ of sub-sigma fields is independent if and only if $$\forall i \in \{1,\ldots,n\}: \Gamma_{i} \in \mathcal{G}_{i} \implies \mathbb{P}\left[\bigcap_{i =1}^n \Lambda_{i}\right]=\prod_{i =1}^n \mathbb{P}\left[\Lambda_{i}\right]$$

*For sub-sigma fields $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$, we denote by $\mathcal{G}_{1} \vee \mathcal{G}_{2}$ the smallest $\sigma$-field that contains $\mathcal{G}_{1} \cup\mathcal{G}_{2}$, i.e., $\mathcal{G}_{1} \vee \mathcal{G}_{2} = \sigma (\mathcal{G}_{1} \cup\mathcal{G}_{2})$.

My question:
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $X$ an integrable random variable. Let $\mathcal D,\mathcal G$ be sub-sigma fields of $\mathcal F$. Assume $\mathcal D$ is independent of $\sigma(X) \vee \mathcal G$.

For $A \in \sigma(X)$, I would like to ask if $\mathbb E [\mathbb{1}_A \mid \mathcal{D} \vee \mathcal{G}]$ is $\mathcal{G}$-measurable.

Please leave me just hints so that I can have a chance to practice. Thank you so much!
 A: Yes, this is true. In fact $\mathbb E [\mathbb{1}_A \mid \mathcal{D} \vee \mathcal{G}]=\mathbb E [\mathbb{1}_A \mid \mathcal{G}]$. To prove this consider the class of all sets of the form $D \cap G$ where $D \in  \mathcal{D}$ and $G \in \mathcal{G}$. This is a $\pi-$ system and it generates  $\mathcal{D} \vee \mathcal{G}$. Since the collection of all sets $E$ such that $\int_E {1}_A dP =\int_E (E[\mathbb{1}_A \mid \mathcal{G})]dP$ is  $\lambda-$ system we only have to verify that this equation holds when $E=D \cap G$. But this is an easy consequence of the independence assumption. 
A: It is clear that $\mathsf{E}[1_{A}\mid \mathcal{G}]=\mathsf{E}[1_{A}\mid \mathcal{G}\vee\mathcal{D}]$ a.s. (by independence). Does it mean that the latter is $\mathcal{G}$-measurable? Take $\mathcal{G}=\{\emptyset,\Omega\}$ and $A$ s.t. $\mathsf{P}(A)=0$. Then the RHS is $0$ but there may be versions of the LHS that are not $\mathcal{G}$-measurable, e.g., $1_B$, where $B\in \mathcal{D}$, $\mathsf{P}(B)=0$, and $B\ne \emptyset$. See this question for more details.
A: I've just completed the proof on the basis of @Kavi Rama Murthy hints. It would be great of someone helps me verify it. Thank you so much!

My attempt:
Let $\mathcal M = \{D \cap G \mid D \in \mathcal D \text{ and } G \in \mathcal G\}$. Then $(\mathcal D \cup \mathcal G) \subseteq \mathcal M \subseteq \sigma (\mathcal D \cup \mathcal G)$. Thus $\sigma(\mathcal M) = \mathcal D \vee \mathcal G$. For $B \in \mathcal M$, we have $B = D \cap G$ for some $(D,G) \in \mathcal D \times \mathcal G$. Consequently, $$\begin{aligned} \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_B] &= \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_{D \cap G}] &&= \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_{D} \mathbb{1}_{G}] &&\overset{(1)}{=} \mathbb E [ \mathbb E [X \mathbb{1}_{G} | \mathcal G]  \mathbb{1}_{D}] \\
&\overset{(2)}{=} \mathbb E [ \mathbb E [X \mathbb{1}_{G} | \mathcal G]] \, \mathbb E[  \mathbb{1}_{D}] &&= \mathbb E [X \mathbb{1}_{G}] \, \mathbb E[  \mathbb{1}_{D}] &&\overset{(3)}{=} \mathbb E [X \mathbb{1}_{G}  \mathbb{1}_{D}]  \\ &= \mathbb E [X \mathbb{1}_{D\cap G}] &&= \mathbb E [X \mathbb{1}_{B}] \end{aligned}$$
(1) follows from: $\mathbb{1}_{G}$ is $\mathcal G$-measurable.
(2) follows from: $\mathbb E [X \mathbb{1}_{G} | \mathcal G]$ is $\mathcal G$-measurable, $\mathbb{1}_{D}$ is $\mathcal D$-measurable, and $\mathcal G$ is independent of $\mathcal D$.
(3) follows from: $X \mathbb{1}_{G}$ is $\sigma(X) \vee \mathcal{G}$-measurable, $\mathbb{1}_{D}$ is $\mathcal D$-measurable, and $\sigma(X) \vee \mathcal{G}$ is independent of $\mathcal D$.
As a result, $\forall B \in \mathcal M: \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_B] = \mathbb E [X \mathbb{1}_{B}]$. Let $\mathcal N = \{B \in \sigma( \mathcal M) \mid \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_B] =  \mathbb E [X \mathbb{1}_B] \} \subseteq \sigma(\mathcal M)$. It follows that $\mathcal M \subseteq \mathcal N$. Clearly, $\mathcal M$ is a $\pi$-system. Next we verify that $\mathcal N$ is a $\lambda$-system:


*

*Clearly, $\Omega \in \mathcal N$.
For $B \in \mathcal N$, $\mathbb E [ \mathbb E [X | \mathcal G]
   \mathbb{1}_{B^c}] = \mathbb E [ \mathbb E [X | \mathcal G](1-
   \mathbb{1}_{B})] = \mathbb E [ \mathbb E [X | \mathcal G]] - \mathbb
   E [ \mathbb E [X | \mathcal G] \mathbb{1}_{B}] = \mathbb E[X] -
   \mathbb E [X \mathbb{1}_B] = \mathbb E [X (1- \mathbb{1}_B)] =
   \mathbb E [X \mathbb{1}_{B^c}]$. Hence $B^c \in \mathcal N$.
Let $(B_n)_{n \in \mathbb N}$ be a sequence of pairwise disjoint
subsets in $\mathcal N$. Then $$\begin{aligned} \mathbb E \left [
   \mathbb E \left [X | \mathcal G \right ] \mathbb{1}_{\bigcup B_n}
   \right ] &= \mathbb E \left [ \mathbb E \left [X | \mathcal G \right
   ] \sum \mathbb{1}_{B_n} \right ] &&= \sum \mathbb E \left [ \mathbb E
   \left [X | \mathcal G \right ]  \mathbb{1}_{B_n} \right ] &&= \sum
   \mathbb E [X \mathbb{1}_{B_n}]\\ &=  \mathbb E \left  [X
   \sum\mathbb{1}_{B_n} \right ]  &&= \mathbb E \left  [X
   \mathbb{1}_{\bigcup B_n} \right ] \end{aligned}$$ Thus ${\bigcup B_n}
   \in \mathcal N$.
By Dynkin's $\pi$-$\lambda$ theorem, we get $\sigma (\mathcal M) \subseteq \mathcal N$ and thus $\sigma (\mathcal M) = \mathcal N$. Hence $\forall B \in \mathcal D \vee \mathcal G: \mathbb E [ \mathbb E [X | \mathcal G] \mathbb{1}_B] = \mathbb E [X \mathbb{1}_{B}]$. Moreover, $\mathbb E [X | \mathcal G]$ is $\mathcal G$-measurable and consequently $(\mathcal D \vee \mathcal G)$-measurable. As a result, $\mathbb E [X | \mathcal G] = \mathbb E [X | \mathcal D \vee \mathcal G]$ almost surely.
