Given three $\sigma$-algebras $\mathcal{A}, \mathcal{B}, \mathcal{C}$ defined on the same space $\Omega$ with $\mathcal{A}$ being independent of both $\mathcal{B}$ and $\mathcal{C}$ is it true that $\mathcal{A}$ is independent of $\sigma(\mathcal{B}, \mathcal{C})$?
This looks like something that must be true but I have neither a proof nor a counterexample.
$\mathcal{B} \cap \mathcal{C} := \{X \subset \Omega \mid X = B \cap C, B \in \mathcal{B}, C \in \mathcal{C} \}$ is a $\pi$-system and it generates $\sigma(\mathcal{B}, \mathcal{C})$ but I don't see how I can show that
$$\mathcal{A} \perp \mathcal{B}, \ \mathcal{A} \perp \mathcal{C} \implies \mathcal{A} \perp \mathcal{B} \cap \mathcal{C}$$
What brought me to this question is the following assertion.
Let $Y_0, Y_1, \ldots$ be iid random variables with $P\{Y_0 = 1\} = P\{Y_0 = -1\} = \frac{1}{2}$. Define $\mathcal{Y} = \sigma(Y_1,Y_2,\ldots), X_n = Y_0Y_1\ldots Y_n$ and $\mathcal{T}_n = \sigma(X_r, r > n)$ for $n = 1,2,\ldots$. Then $Y_0$ is independent of $\sigma(\mathcal{Y}, \cap_n \mathcal{T}_n)$.
Obviously, $Y_0$ is independent of $\mathcal{Y}$. It is also not hard to show that it is independent of $\mathcal{T}_1$, hence of the tail $\sigma$-algebra. Here I could note that $\cap_n \mathcal{T}_n$ contains either null-measure or full-measure events since $X_1,X_2,\ldots$ is an independent sequence, which I can also show. Then defining the $\pi$-system $\mathcal{Z} := \mathcal{Y} \bigcap \cap_n \mathcal{T}_n$ I could write
$$P\{\{Y_0 = 1\} \cap Z \} = P\{\{Y_0 = 1\} \cap (Y \cap T)\}$$ for some $\mathcal{Z} \ni Z = Y \cap T$ with $Y \in \mathcal{Y}$ and $T \in \cap_n \mathcal{T}_n$. Continuing the expression above
$$P\{\{Y_0 = 1\} \cap (Y \cap T)\} = P\{\left(\{Y_0 = 1\} \cap Y\right) \cap T)\}= P\{Y_0 = 1\} P\{Y \cap T\}$$
I did not show all the steps above but I considered the cases $P\{T\} = 1$ and $P\{T\} = 0$ separately. So this proves independence of $Y_0$ and $\sigma(\mathcal{Y}, \cap_n \mathcal{T}_n)$. I was hoping (for whatever reason) that this would be true regardless of $\cap_n \mathcal{T}_n$ being a trivial $\sigma$-algebra.
If my first claim is not true, is there a result that gives necessary and sufficient conditions under which it is true? A counterexample would also be appreciated.
The answer to your question is: no.
Let $X,Y,Z$ be random variables on the same probability space.
It can happen that random variables $X$ and $Y$ are independent and also $X$ and $Z$ are independent, but that not $X$ and $(Y,Z)$ are independent.
E.g. let $X$ and $Y$ be iid that both only take values in $\{0,1\}$ with $P(X=1)=P(Y=1)=0.5$ and let $Z$ take value $1$ if $X=Y$ and let $Z$ take value $0$ otherwise.
This can be modeled like this:
Let $\Omega=\Omega_1\cup\Omega_2\cup\Omega_3\cup\Omega_4$ where the $\Omega_i$ are non-empty and disjoint.
Let $\mathcal D=\sigma(\{\Omega_i\mid i=1,2,3,4\})$ and let $P$ be a probability measure on $(\Omega,\mathcal D)$ that is determined by $P(\Omega_i)=\frac14$ for $i=1,2,3,4$.
Let $X=1_{\Omega_1}+1_{\Omega_2}$, $Y=1_{\Omega_1}+1_{\Omega_3}$ and $Z=1_{\Omega_1}+1_{\Omega_4}$.
Then it can be proved that $X$ and $Y$ are independent and that $X$ and $Z$ are independent.
Setting $\mathcal A=\sigma(X)$, $\mathcal B=\sigma(Y)$ and $\mathcal C=\sigma(Z)$ that means that $\mathcal A$ is independent of both $\mathcal B$ and $\mathcal C$.
However $\mathcal A\subseteq\mathcal D=\sigma(\mathcal B,\mathcal C)$ so $\mathcal A$ and $\sigma(\mathcal B,\mathcal C)$ are not independent.