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Answer by Elliot Glazer for Extending the product measure on $2^\omega$

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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.


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