Question: For $n\ge 3$, is there a maximal isometrically-invariant finitely additive extension of Lebesgue measure on $\mathbb R^3$?

A maximal extension of a measure with some properties (in this case, isometric invariance and finite additivity) is one that cannot be extended to a larger algebra of subsets while preserving these properties.

Ciesielski and Pelc (see here) proved that there is no maximal isometrically-invariant countably additive extension of Lebesgue measure on $\mathbb R^n$. On the other hand, Banach showed that for $n\le 2$, there is a finitely additive extension of Lebesgue measure to all subsets of $\mathbb R^n$, and of course it follows from Banach-Tarski that this is false for $n\ge 3$.

I am assuming Choice throughout, of course.

Question: For $n\ge 3$, is there a maximal isometrically-invariant finitely additive extension of Lebesgue measure on $\mathbb R^3$?

A maximal extension of a measure with some properties (in this case, isometric invariance and finite additivity) is one that cannot be extended to a larger algebra of subsets while preserving these properties.

Ciesielski and Pelc (see here) proved that there is no maximal isometrically-invariant countably additive extension of Lebesgue measure on $\mathbb R^n$. On the other hand, Banach showed that for $n\le 2$, there is a finitely additive extension of Lebesgue measure to all subsets of $\mathbb R^n$, and of course it follows from Banach-Tarski that this is false for $n\ge 3$.

I am assuming Choice throughout, of course.

[Edit: As Bill Johnson points out in the comments, this is just a quick application of Zorn. I guess that since this is very much nontrivial in the countably additive case, it didn't occur to me that maybe the finitely additive case is really easy!]