If $A_1,A_2$ are independent, are $A_1\cap B, A_2\cap B$ independent? Let $A_1,A_2$ be independent events. Can we say that $A_1\cap B, A_2\cap B$ are also independent for any other event $B$?
After a few attempts of proving the affirmative, I'm convinced that the answer is negative.
But how should I understand this intuitively? If two events are independent, shouldn't their restrictions to a subspace be independent as well?
Any help is appreciated. Thank you!
 A: Conceptually, independence defines a strict relationship of the intersection of two events the to the events themselves. However, we have a lot more latitude in what $A_1 \cap B$ and $A_2\cap B$ can be by different choices of $B$ so there is not prima facie reason to think this would hold.
Assuming the above intersections are non-empty, for these to be independent, we need:
$$(*)\;\;\frac{P(A_1\cap A_2 \cap B)}{P(A_2 \cap B)} = P(A_1\cap B),\;\;\forall B $$
Let's define a sequence of subsets of $A_1 \cup A_2$ where we've taken "bites" out of $A_2$ but not $A_1$. Each of these is a possible set $B$ in above and we'll consider the conditional probabilities.
Let $C_1 \subset C_2 \subset C_3 ...$ be an increasing sequence of sets such that $\bigcup_{i=1}^{\infty} C_i = A_2\setminus A_1$ and  $B_i := A_1 \cup A_2\setminus C_i$. Note that $A_1 \cap B_i = A_1 \cap A_2$ and $A_2 \cap B_i \neq \emptyset\;\;\forall i> 0$
Then
$$P(A_1\cap B_i|A_2\cap B_i) = \frac{P(A_1 \cap A_2 \cap B_i)}{P(A_2 \cap B_i)}= \frac{P(A_2 \cap (A_1 \cap B_i))}{P(A_2 \setminus C_i)} = $$
$$\frac{P(A_2 \cap A_1) }{P(A_2 \setminus C_i)}=\frac{P(A_2)P(A_1) }{P(A_2 \setminus C_i)}\text{    (by independence of $A_1$ and $A_2$)}$$
If $(*)$ is true,
$$P(A_1\cap B_i|A_2\cap B_i) = P(A_1\cap B_i)=P(A_1) \;\;\forall i>0$$
However, that requires
$$P(A_2\setminus C_i) = P(A_2) \implies C_i = \emptyset$$
Since $C_i$ is increasing, that means that independence only holds when $B = A_1 \cup A_2$. Otherwise,
$$P(A_2\setminus C_i) < P(A_2) \implies P(A_1\cap B_i|A_2\cap B_i) > P(A_1)\;\quad\square$$

Some additional details
Looking at the limiting behavior as of $B_i$ under the increasing sequence $C_i$ (above) we see
$$\lim_{i \to \infty} \frac{P(A_2 \cap A_1) }{P(A_2 \setminus C_i)} = \frac{P(A_2 \cap A_1) }{P(A_2 \setminus (A_2\setminus A_1))} = \frac{P(A_2 \cap A_1) }{P(A_2 \cap (A_2 \cap A_1^c)^c)}=$$
$$\frac{P(A_2 \cap A_1) }{P(A_2 \cap (A_2^c \cup A_1))} = \frac{P(A_2 \cap A_1) }{P(A_1\cap A_2)} = 1 \neq P(A_1)$$
Similarly, let $D_1 \supset D_2 \supset D_3 ...$ be a decreasing sequence of sets such that $\bigcap_{i=1}^{\infty} D_i = \emptyset$ and  $B_i := A_1 \cup A_2\setminus D_i$, then
$$\frac{P(A_1 \cap A_2 \cap B_i)}{P(A_2 \cap B_i)} = \frac{P(A_2 \cap A_1) }{P(A_2 \setminus D_i)} \implies$$
$$\lim_{i \to \infty} \frac{P(A_2 \cap A_1) }{P(A_2 \setminus D_i)} = \frac{P(A_2 \cap A_1) }{P(A_2 \setminus \emptyset)} = P(A_1|A_2) = P(A_1)\text{ by independence}$$
Therefore, depending on our choice of $B$, we can make $A_1\cap B$ and $A_2 \cap B$ as dependent or independent as we want.
A: A very simple counterexample is $B=A_2$. Then $B\cap A_2=A_2$ and $B\cap A_1=A_2\cap A_1$ and clearly $p(A_1\cap B)$ is not independent of $A_2$.
