Does independence of the sequence $f(A_i, B)$ imply the sequence is independent of $B$? Suppose $B, \{A_i: i \in \omega\}$ are i.i.d. random variables with uniform distributions on $[0,1]$. If $f$ is a map such that $\{f(A_i, B): i \in \omega\}$ are independent, must $\{f(A_i, B): i \in \omega\}$ also be independent of $B$? 
 A: It is true.
Let $\chi_Z(x,y)$ be the indicator function of the set $\lbrace (x,y) \,|\, f(x,y)\in Z\rbrace$ for some set $Z$.
The condition that the events $f(A_1,B)\in Z$ and $f(A_2,B)\in Z$ are independent is that
$$ \int_{[0,1]^3} \chi_Z(x,y)\chi_Z(x'y)\,dxdx'dy 
   = \int_{[0,1]^2} \chi_Z(x,y) \,dxdy \ \ \int_{[0,1]^2} \chi_Z(x',y') \,dx'dy'.$$
Define $g(y) = \int_0^1 \chi_Z(x,y)\,dx$.  Then the above condition can be rearranged into
$$ \int_{[0,1]^2} g(y) (g(y)-g(y'))\, dy dy' = 0.$$
That's just a change of variables from 
$$ \int_{[0,1]^2} g(y') (g(y)-g(y'))\, dy dy' = 0,$$
so we can subtract the two to get
$$ \int_{[0,1]^2} (g(y)-g(y'))^2 \,dydy' = 0$$
which by positivity means $g(y)$ is constant a.e.. Harking back to the definition of $\chi_Z$ and noting that the set $Z$ was arbitrary, we see that $f(A_1,B)$ is independent of $B$.
I have only considered pairwise independence, but once you find that $g(y)$ is independent of $y$ it makes the whole lot of them independent.  At least, it seems so at 3am...
ADDED after waking up: At the moment I can't see how uniform distributions are needed for this argument. Just consider any vertical measure $\mu$ and horizontal measure $\nu$ and replace $dx$ by $d\mu$, $dy$ by $d\nu$, etc, in the above argument. Still works, doesn't it?
MORE ADDED: Nate's question leads me to note the following, which  surely must be well-known in the theory of exchangeable random variables (but I didn't know it).
Theorem. Let $X,Y$ be random variables such that $(X,Y)$ and $(Y,X)$ have the same distribution.  Then $X$ and $Y$ are independent iff for all measurable sets $S$, $P(X\in S \wedge Y\in S)=P(X\in S)^2$.
Proof. Let $S,T$ be measurable sets. First assume $S$ and $T$ are disjoint.  Since by the exchangeability assumption $P(X\in T\wedge Y\in S)=P(X\in S\wedge Y\in T)$, we have
\begin{align}
  2 P(X\in S\wedge Y\in T)&=P(X,Y\in S\cup T) - P(X,Y\in S) - P(X,Y\in T) \\
   &= P(X\in S\cup T)^2 - P(X\in S)^2 - P(X\in T)^2 \\
   &= (P(X\in S) + P(X\in T))^2 - P(X\in S)^2 - P(X\in T)^2 \\
   &= 2 P(X\in S) P(X\in T) \\
   &= 2 P(X\in S) P(Y\in T).
\end{align}
If $S$ and $T$ are not disjoint, break $P(X\in S\wedge Y\in T)$ into four disjoint cases like $P(X\in S\wedge Y\in T\setminus S)$ and apply the above to each.  It works.
