I want to prove the following lemma:
Let $(\Omega, \mathcal{F}, P)$ be a probability space. Suppose that $\mathcal{G}, \mathcal{H}, \mathcal{I}$ are sub-algebras of $\mathcal{F}$. Then we have that for all $A \in \mathcal{G}$, $B \in \mathcal{H}$, with probability $1$, $$P(A|\mathcal{I} \vee \mathcal{H}) = \frac{P(A \cap B | \mathcal{I})}{P(B | \mathcal{I})}$$
We use the definition that $P(A|\mathcal{G}) = E(\mathbb{1}_A|\mathcal{G})$, where $E(X|\mathcal{A})$ denotes any $\mathcal{A}$-measurable set $Z$ such that $E(\mathbb{1}_XA) = E(\mathbb{1}_ZA)$ for all $A \in \mathcal{A}$. Such $Z$ is unique up to a probability $0$ set.
Here is my attempt: Fix $A \in \mathcal{G}$. By definition of conditional expectation, we can find a $\mathcal{I} \vee \mathcal{H}$-measurable random variable $Z^{'} = P(A|\mathcal{I} \vee \mathcal{H}) = \mathbf{E}(\mathbb{1}_A | \mathcal{I} \vee \mathcal{H})$ such that for any $B^{'} \in \mathcal{I} \vee \mathcal{H}$, we have that $\mathbf{E}(Z^{'} \mathbb{1}_{B^{'}} ) = \mathbf{E}(\mathbb{1}_A \mathbb{1}_{B^{'}} )$ with probability $1$.
Note that for any $\mathcal{H}$-measurable set $B$, we have that $\mathbf{E}[(Z^{'}-\mathbb{1}_A) \mathbb{1}_B ] = 0 $ by definition of $Z^{'}$. Moreover, $\mathbf{E}[(Z^{'}-\mathbb{1}_A) \mathbb{1}_B | \mathcal{I}] = 0 $ as well, because by definition of conditional expectation $\int_D Z^{'} d\mu$ and $\int_D A d\mu$ agree on all sets $D \in \mathcal{I} \vee \mathcal{H}$, including those only in $\mathcal{I}$. We thus have that $ \mathbf{E}(Z^{'} \mathbb{1}_B | \mathcal{I}) = \mathbf{E} [\mathbb{1}_A \mathbb{1}_B | \mathcal{I}] $.
Now, define a new measure $\nu$ on the sigma algebra $\mathcal{I}$ by $\nu(A) = \int_A Z^{'} d\mu $, which is absolutely continuous to $\mu$. Taking the Radon Nikodym derivative of $\nu$, we can obtain $d\nu = \mathbb{1}_Z d\mu$ for certain $\mathcal{I}$-measurable set $Z$. It is obvious that $\mathbb{1}_Z$ and $ \mathbb{1}_{Z^\prime}$ agrees on all $\mathcal{I}$-measurable sets, and they are both $\mathcal{I} \vee \mathcal{H}$-measurable. Thus $Z = Z^{'}$ with probability $1$. We also have that $\mathbf{E}[Z^{'} \mathbb{1}_B | \mathcal{I}] = Z \mathbf{E}[ \mathbb{1}_B | \mathcal{I}]$. Since $Z \in \mathcal{I}$, we have $Z \mathbf{E}(\mathbb{1}_B | \mathcal{I}) = \mathbf{E}(\mathbb{1}_A \mathbb{1}_B | \mathcal{I})$ and thus $Z^{'} = Z = \frac{\mathbf{E}(\mathbb{1}_A \mathbb{1}_B | \mathcal{I})}{\mathbf{E}(\mathbb{1}_B | \mathcal{I})}$ with probability $1$. This completes the proof.
However, I am not sure if the proof above is correct. Can someone point me any sources in which proper proof is given? Thank you very much!