Why are these two sigma algebras independent? Given a probability space, let $E$ be an event and let $\{E_n\}_{n=1}^\infty $ be a sequence of events.
Claim: If $\sigma(E)$ is independent of $\sigma(E_n)$ for each $n$, then $\sigma(E)$ is independent of $\sigma(E_1,E_2,\dots)$.
My attempt was to show that $\{ E \} \cup \{ E_n\}$ is mutually independent, but the assumption of the claim just gives that $\{ E \} \cup \{ E_n\}$ is pairwise independent, which does not imply mutual independence.
How should I approach to prove the claim?
 A: The claim is false. Consider the probability space $\Omega = \{0, 1\}^2$ with the uniform distribution and the events $E = \{(0, 0), (1, 1)\}, E_1 = \{(0, 0), (0, 1)\}$ and $E_2 = \{(0, 0), (1, 0)\}$ and $E_k = \emptyset$ for $k > 2$. Note that $\sigma(E_1, E_2) = \mathcal{P}(\{0, 1\}^2)$.
The interpretation of this example is as follows: Consider two independent coins and the random Variables $X_1, X_2$ that represent the result of the first respectively second coin. Define $X_3$ to be 1 if $X_1 = X_2$ and 0 else. Then $X_1$ and $X_3$ are independent, $X_2$ and $X_3$ are independent, but $X_3$ and $(X_1, X_2)$ are not independent.
However, the initial claim is true if one demands the independence of $\sigma(E)$ and $\sigma(E_1, \ldots, E_n)$ for all $n$.
A: Try showing this.  Suppose your probability space is $(\Omega, \mathcal{S}, P)$.
If $E$ is an event, 
$$\mathcal{F} = \{Q\in \mathcal{S}| P(E\cap Q) = P(E)\cdot P(Q)\}$$
is a $\sigma$-subalgebra of $\mathcal{S}$.
Your result would follow immediately from this.
