It is known that there exists a finitely additive translation invariant measure on $\mathbb{R}$ that extends the Lebesgue measure. I.e. a function $m:\mathcal{P}(\mathbb{R}) \rightarrow [0,\infty]$ that is

  1. Finitely additive: $m(A \sqcup B) = m(A) + m(B)$

  2. For a Lebesgue measurable set $A$, $m(A)$ is its Lebesgue measure.

  3. For all $A \subset \mathbb{R}$, $\lambda \in \mathbb{R}$ we have $m(A + \lambda) = m(A)$

The proof can be found in Stein & Shakarachi's Functional Analysis (the main tool is the Hahn Banach Theorem, which gives the "Banach Integral"). This got me wondering: can we also find such an m that is homogeneous? Meaning $\forall \lambda \in \mathbb{R}, \, A \subset \mathbb{R} \; m(\lambda A) = \vert \lambda \vert m(A)$. I couldn't see directly from the proof how to do so. Does such a measure even exist? Any reference given is appreciated.

A similar statement in $\mathbb{R}^3$ fails because of the Banach-Tarski Paradox.


[Corrected answer, entirely rewritten] Yes. And this also works on $V=\mathbf{R}^n$ (with homogeneity axiom rewritten as $\mu(tY)=|t|^n\mu(Y)$).

Let $\mu_0:\mathcal{P}(V)\to [0,\infty]$ be a translation invariant, finitely additive mesure, extending the Lebesgue measure.

Let $I$ be the set of $Y\subset V$ such that $$\sup_{t\in\mathbf{R}^*}|t|^{-n}\mu_0(tY)<\infty.$$ This is an ideal of the Boolean algebra $\mathcal{P}(V)$. For $Y\in\mathcal{P}(V)\smallsetminus I$, define $\mu(Y)=\infty$.

Let $\nu$ be an invariant mean on the discrete (amenable) group $\mathbf{R}^*$. For $Y\in I$, define $$\mu(Y)=\int_{t\in\mathbf{R}^*}|t|^{-n}\mu(tY)d\nu(t).$$ (This is valid since we integrate a bounded function along the mean.) Then $\mu$ is a finitely additive, translation-invariant measure on $I$, and hence on $\mathcal{P}(V)$ since $I$ is an ideal. From the invariance of $\nu$, we deduce that $\mu$ satisfies $\mu(tY)=|t|^n\mu(Y)$ for all $t\in\mathbf{R}^*$.

  • $\begingroup$ Thank you for the answer! Could you please briefly explain what you mean by a mean on $\mathcal{P}(V)$? As it seems it can't obtain the value $\infty$, so how can it extend the Lebesgue measure? ($m(\mathbb{R}) = \infty$). $\endgroup$ – pitariver Feb 23 at 19:20
  • $\begingroup$ @pitariver you're right, my answer is not correct. In a first thought, I can repair the proof so as to work on bounded subsets. I have to think more... $\endgroup$ – YCor Feb 23 at 19:27
  • $\begingroup$ I think if your approach proves the existence of such mean for $\mathbb{R}^n / \mathbb{Z}^n$ it solves the problem, since form there we can manually extend the mean to $\mathbb{R}^n$. In the case of $\mathbb{R}$ one can define $$ m(E) = \sum_{k=-\infty}^{\infty} \hat{m}(E \cap [k,k+1) - k) $$ where $\hat{m}$ is the promised mean on $\mathbb{R} / \mathbb{Z}$. The proof in the book shows that it works. $\endgroup$ – pitariver Feb 23 at 19:35
  • $\begingroup$ I rewrote the answer using another approach. $\endgroup$ – YCor Feb 23 at 19:43
  • $\begingroup$ I think you got it, thanks again for the attention! $\endgroup$ – pitariver Feb 23 at 20:05

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