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(Cross posted at MSE.)

The answer to this question shows that there are translation-invariant extensions of Lebesgue measure.

Are there maximal translation-invariant extensions of Lebesgue measure (in ZFC)? If so, is there more than one?

To be more precise, let $\mathcal L$ be the sigma-algebra of Lebesgue measurable subsets of $[0,1]$. By the linked post, there exists $A \subset [0,1]$ such that $A \notin \mathcal L$ and there exists a translation-invariant measure $\mu'$ on $\sigma(\mathcal L \cup \{A\})$ that extends the Lebesgue measure on $\mathcal L$. My question is:

Is there a sigma-algebra $\mathcal L' \supset \mathcal L$ and a measure $\mu'$ on $\mathcal L'$ such that $\mu'$ is a translation-invariant extension of Lebesgue measure and if $A \not \in \mathcal L'$, then there is no translation-invariant extension of $\mu'$ to $\sigma(\mathcal L' \cup \{A\})$? If such an $\mathcal L'$ exists, is it unique?

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The answer is no. This is the main result of "Extensions of invariant measures on Euclidean spaces" by Ciesielski and Pelc.

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  • $\begingroup$ Cool paper, thanks! $\endgroup$
    – aduh
    Commented Jan 27, 2021 at 8:49
  • $\begingroup$ Is there a finitely additive isomery invariant extension of Lebesgue measure(in $\mathbb{R}^d$) to a maximal sigma algebra? $\endgroup$
    – Vivaan Daga
    Commented Jul 8, 2023 at 18:59
  • $\begingroup$ @VivaanDaga Yes, by Zorn's lemma. $\endgroup$
    – Elliot Glazer
    Commented Jul 8, 2023 at 19:53
  • $\begingroup$ @ElliotGlazer We get an algebra by Zorn's lemma, not a sigma algebra unless I am missing something. $\endgroup$
    – Vivaan Daga
    Commented Jul 8, 2023 at 20:03
  • $\begingroup$ Or is that an open problem?(It feels like a hard question) $\endgroup$
    – Vivaan Daga
    Commented Jul 9, 2023 at 5:51

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