Finitely additive measures on $\mathbb Z_2^\omega$ with invariance and independence constraints Let $G = \mathbb Z_2^\omega$, with pointwise addition.  Assume the Axiom of Choice.  I am interested in finitely additive probability measures $\mu$ defined on all of $\mathcal PG$ that can be intuitively thought to represent an infinite sequence of independent fair coin tosses.  
One condition I want is $G$-invariance: $\mu(A)=\mu(\alpha+A)$ for $\alpha\in G$ and $A\subseteq G$.  Since $G$ is abelian, there is an invariant measure on $\mathcal PG$.
Question: Is there a finitely additive invariant $\mu$ also subject to the independence condition that $\mu(A\cap B)=\mu(A)\mu(B)$ whenever $A$ and $B$ "depend on different coordinates".  
(If we let $\pi_J : G\to \mathbb Z_2^J$ be the projection for $J\subseteq \omega$ given by $\pi_J(\alpha)=\alpha|_J$, then $A$ and $B$ depend on different coordinates iff we can write $A=\pi_J^{-1}[A']$ and $B=\pi_K^{-1}[B']$ for some disjoint $J$ and $K$.)
Comment: I was asked if there are $G$-invariant $\mu$ that don't satisfy independence.  Yes.  Start by noting that there is no $G$-invariant $\mu$ that is also invariant under the permutations of $\omega$ (acting by composition--I will write this action on the right).  In fact, for any $G$-invariant $\mu$ on $G$ and infinite $A\subseteq \omega$ there will be a permutation $\tau$ fixing $\omega-A$ and a subset $C$ of $G$ depending only on the coordinates in $A$ such that $\mu(C)\ne\mu(C\tau)$.  
Now, fix any $G$-invariant $\mu$, and let $A$ and $B$ be infinite disjoint subsets of $\omega$.  It's easy to now show that there is a $G$-invariant $\nu$ (formed by combining $\mu$ with two permutations) and subsets $C$ and $D$ of $G$ depending on the coordinates in $A$ and $B$ respectively such that $\nu(C)\ne \mu(C)$ and $\nu(D) \ne \mu(D)$. Suppose $\mu$ and $\nu$ satisfy the independence condition.  Replacing $C$ and/or $D$ with its complement as needed, assume $\nu(C)>\mu(C)$ and $\nu(D)>\mu(D)$.  Let $\rho=(1/2)(\mu+\nu)$.  Then $\rho$ is $G$-invariant but  $\rho(C\cap D) > \rho(C)\rho(D)$, so $\rho$ does not satisfy the independence condition.
 A: It's been a while since I've worked with amenable groups and integrals against finitely additive measures, so I could be missing something, but it now seems to me that the question is easy. While apparently in general the direct product of amenable groups isn't amenable, the direct product of abelian groups is of course amenable. So we can use:
Theorem. If $G$ is the direct product of the groups $(G_i)_{i\in I}$ and $G$ is amenable, then there is a finitely additive invariant measure on $G$ that has independence for finite sequences of events depending respectively on pairwise disjoint sets of coordinates.
We need only prove the Theorem for finite sets $I$. For let $F$ be the set of finite partitions of $I$, with a partial
order given by $K\le L$ iff $L$ is at least as fine as $K$.
For any partition $K\in F$, we consider $G$ to be the product of the $G^A=\prod_{i\in A} G_i$ as $A$ ranges over $K$. The measure $\mu_K$ we get from the finite index-set case of the theorem applied to the product of the $G^A$ will satisfy the independence condition for finite sequences of events depending on pairwise disjoint sets of coordinates that are in the algebra generated by $K$.
Let $\mu$ be a limit point of $\mu_K$ along the net $(F,\le)$. Then $\mu$ is the requisite measure on $G$.
The finite index-set case follows by iterating:
Lemma: If $G_1$ and $G_2$ are groups with finitely additive invariant measures $\mu_1$ and $\mu_2$, then there is
a finitely addditive invariant measure $\mu$ on their direct product such that $\mu(A\times B)=\mu_1(A)\mu_2(B)$.
Proof of Lemma: Let $\mu(U) = \int_{G_2} \mu_1 ( \{ x_2 : (x_1,x_2) \in U \} ) d\mu_2(x_2)$. Integrals against f.a. measures are additive so $\mu$ is a f.a. measure, and invariance is obvious.
