Extending the product measure on $2^\omega$

Consider the standard completed product measure $P$ on $\Omega=\{0,1\}^\omega$ corresponding to an i.i.d. sequence of fair coin-flips.

Given $n\in\omega$, let $\rho_n$ be the bijection of $\Omega$ to itself given by flipping the value of the $n$th coordinate.

Say that a subset $A$ of $\Omega$ has property $H$ iff there is an $n$ such that $\rho_n(A)\cap A=\varnothing$ and $\rho_n(A)\cup A=\Omega$.

Assume AC.

Question: What can we say about the existence of (countably-additive) extensions of $P$ that assign measure $1/2$ to all subsets with property $H$? Are there any? If so, are there any with nice invariance properties, like invariance under $\rho_n$ and/or under bijections of $\{0,1\}^\omega$ induced by permutations of $\omega$?

Remarks: Intuitively, the probability of $A$, if $A$ has property $H$, should be $1/2$. This is true with respect to $P$ if $A$ is $P$-measurable. But given AC, there are $P$-nonmeasurable $A$ having $H$. Say that $\alpha\sim\beta$ iff one can get $\beta$ from $\alpha$ by applying an even number of the $\rho_n$. Given a $\sim$-equivalence class $U$, let $U' = \rho_1 U$. Now take a set $B$ of equivalence classes that contains exactly one member of $\{ U,U' \}$ for every $U$, and let $A$ be the union of the members of $B$. Then $A$ has $H$ with respect to every index $n$, and hence is non-measurable by Lévy's zero–one law.

• Are you sure about your construction of a non-measurable set $A$ with property $H$? $A$ contains exactly one binary sequence with a finite even number of $1$s and exactly one binary sequence with a finite odd number of $1$s (call this number $j$). Then for any $n$, $\rho_n(A)$ contains exactly one binary sequence with a finite even number of $1$s, and that number is either $j+1$ or $j-1$. So $A\cup \rho_n(A)$ contains at most $2$ sequences with a finite even number of $1$s, and the union can't be all of $\Omega$. Sep 11, 2017 at 15:59
• But I think you can still get a nonmeasurable subset of $\Omega$ with property $H$ by fixing a non-measurable subset $X\subseteq \{0,1\}^{\omega_+}$ and letting $A = \{0\eta\mid \eta\in X\}\cup \{1\eta\mid \eta\notin X\}$. Sep 11, 2017 at 16:00
• Sorry, I had forgotten how the construction went (I came up with it a couple of years ago). I now edited to include the correct construction, but yours is neater and more illuminating, I think. Sep 11, 2017 at 16:16

The answer to the main question is negative given the continuum hypothesis (or more generally the non-existence of a real valued measurable cardinal less than or equal to $\mathfrak c$). An extension $\mu$ of $P$ to all sets with property $H$ (even without any invariance properties) gives rise to an atomless probability measure $\nu$ on all subsets of $2^\omega$, the existence of which was shown to contradict the CH by Banach and Kuratowski.

Here's how we get $\nu$. Given any subset $A$ of $2^\omega$, let $A^* = (\{ 0 \} \times A) \cup (\{ 1 \}\times (2^\omega\backslash A))$. This has property $H$. Thus, $\{ 0 \} \times A = A^* \cap (\{ 0 \} \times 2^\omega)$ is $\mu$-measurable, and so we can define $\nu(A) = 2\mu(A \cap (\{ 0 \} \times 2^\omega))$, which will be a probability measure on $\mathcal P {2^\omega}$, and it will be atomless because $P$ is. (This uses Alex Kruckman's simple construction of sets satisfying $H$.)

Given ZFC alone, one can give a negative answer to one of my follow-ups. Specifically, there is no extension of the product measure to all sets with property $H$ that is invariant under permutations of $\omega$, even if we just ask for finite additivity in the extension. For if $\mu$ is a finitely additive extension of $P$ to all sets with $H$, then $\nu$ (defined as above) will be a finitely additive probability measure on $\mathcal P {2^\omega}$ invariant under permutations. A contradiction follows from the fact that $F_2$ is a subgroup of $S_\omega$. More precisely: Let $\langle Q_n \rangle_{n\in\omega}$ be a partition of $\omega$ into countably infinite subsets. Let $\phi_n$ be a bijection of $\omega$ with $Q_n$, and let $X$ be the subset of $2^\omega$ consisting of functions $f:\omega\to \{0,1\}$ such that $f\circ \phi_n \ne f\circ \phi_m$ if $n\ne m$. Then $P(X)=1$ (the probability of a repeat in a countable sequence of uniform i.i.d. random variables is zero).

Now for any permutation $\pi$ of $\omega$, let $\pi^*$ be the permutation such that $\pi^*(\phi_n(m))=\pi^*(\phi_{\pi(n)}(m))$ for all $n$ and $m$. Note that $\pi^*[X]=X$. The permutations of the form $\pi^*$ form a subgroup $G$ of $S_\omega$ isomorphic to the full group, and $G$ has no non-trivial fixed points on $X$. Since $S_\omega$ contains a free group of rank 2, there is no finitely additive $G$-invariant probability measure on $X$ (here we use AC), which contradicts the fact that $\nu$ is such a measure.

I still don't know: Can one prove a negative answer to the main question in ZFC alone? Can one prove a negative answer to the main question with $\rho_n$-invariance in ZFC alone?

There is no such measure. Suppose toward contradiction $$\mu$$ is such a measure.

Partition $$\Omega$$ by the mod finite equivalence relation $$\sim_{\text{fin}},$$ let $$X \subset \Omega$$ be a choice of representatives and $$\pi: \Omega \rightarrow X$$ send each sequence to its representative.

Let $$X_n = \{x \in \Omega: x \triangle \pi(x) \subset [n]\}.$$ Each $$X_n$$ can be extended to some $$H_n$$ satisfying $$\Omega \setminus H_n = \rho_{n+1}(H_n),$$ and therefore having property $$H.$$ For each $$x \in X,$$ there are cofinitely many $$n$$ such that $$x \in X_n \subset H_n.$$ Thus, $$\Omega = \bigcup_{n \in \mathbb{N}} \bigcap_{m \ge n} H_m.$$ This is an increasing union of sets of $$\mu$$-measure bounded by $$\frac{1}{2},$$ contradiction.