$\mathcal A \perp \bigvee _i \mathcal B_i \Longleftrightarrow \forall i\quad \mathcal A \perp \mathcal B_i$ Let $(\Omega, \mathcal F , \mathbb P)$ be a probability space, $(\mathcal A_i)_{i \in I}$ a family of sigma-algebras of this space and $\mathcal B$ another sigma-algebra of this space. Then I claim that
$$
\mathcal A \perp \bigvee _i \mathcal B_i \Longleftrightarrow \forall i~~\mathcal A \perp \mathcal B_i
$$
where $\perp$ stands for the independence and $\vee_i \mathcal B_i$ is the supremum sigma-algebra:
$$
\bigvee _i \mathcal B_i := \sigma \left( \bigcup_i \mathcal B_i \right).
$$
Here goes my proof, I would like to know if there is any mistake, as I have never seen this result in any of my references I am not sure it is correct. Any comments or feedback are welcome.
$\Longrightarrow$ : If we assume that $\mathcal A \perp \bigvee _i \mathcal B_i$ then we have forall $j \in I$
$$
\mathcal B_j \subset \bigvee _i \mathcal B_i 
$$
so $\mathcal B_j \perp \mathcal A$.
$\Longleftarrow$ : Assume that all the $\mathcal B_i$ are independant of $\mathcal A$. Then consider
$$
\mathcal C = \{ B \in \vee_i \mathcal B_i : B \perp \mathcal A \} = \{ B \in \vee_i \mathcal B_i : \forall A \in \mathcal A,~~B \perp  A \},
$$
by hypothesis we have that any $\mathcal B_i$ is contained in $\mathcal C$. To conclude the proof we are left to show that $\mathcal C$ is a sigma-algebra:

*

*The empty set is independant of any sigma-algebra and $\vee_i \mathcal B_i$ being a sigma-algebra, $\emptyset \in \mathcal C$

*If $B \in \mathcal C$ then $B^c \in \vee_i \mathcal B_i$ and we also have $B^c \perp \mathcal A$ so $B^c \in \mathcal C$

*If the $B_n$ are in $\mathcal C$ and disjoints, for any $A \in \mathcal A$ we have
$$
\mathbb P \left[ A \cap \left( \bigcup_n B_n \right) \right] = \mathbb P \left[ \bigcup_n A \cap B_n  \right] = \sum_n  \mathbb P (A \cap B_n)= \sum_n \mathbb P(A)\mathbb P(B_n)
$$
and factorizing the right hand side to $\mathbb P(A) \mathbb P(\cup_n B_n)$ is legal because the series $\sum \mathbb P (B_n)$ converges.

Edit: As suggested in the comments of the answer my mistake was to replace the countable union by disjoint countable union in the definition of a sigma-algebra. Moreover this result is not true, see in the answer.
 A: Your proof that $\mathcal A \perp \bigvee_i \mathcal B_i$ implies $\mathcal A \perp \mathcal B_j$ for all $j$ is correct, but it is not true that $\mathcal A \perp \mathcal B_j$ for all $j$ implies $\mathcal A \perp \bigvee_i \mathcal B_i$.  I agree with your proof that $\mathcal C$ is a $\sigma$-algebra and that $\mathcal C \perp \mathcal A$, but this doesn't imply $\mathcal A \perp \bigvee_i \mathcal B_i$.
For example, we could consider i.i.d. random variables $X_1, X_2$ with $\mathbb{P}(X_1 = 1) = \mathbb{P}(X_1 = -1) = \frac 12$, and define $Y := X_1 X_2$.  Let $\mathcal A = \sigma(Y)$ and $\mathcal B_j = \sigma(X_j)$.  Then $\mathcal A \perp \mathcal B_j$ for all $j$, but $\mathcal A$ is not independent of $\bigvee_i \mathcal B_i$, and in fact $\mathcal A \subset \bigvee_i \mathcal B_i$.
A: Your proof is wrong (I'm not sure that the theorem is even true), though you are on the right track. You haven't shown that $C$ is a $\sigma$-algebra because you need closure under all countable unions, not just disjoint ones. What you have shown is that $C$ is a lambda-system. The standard tool for extending results to sigma algebras from pi systems is the pi-lambda theorem. If you modify your approach a little, you can prove a version of the theorem you seek. See the chapter on independence in Klenke's probability theory book for detailed proofs using this technique (a very robust technique!).
