# Finitely Additive Homogeneous Translation Invariant Measure on $\mathcal{P}(\mathbb{R})$

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.

• I think that, to run afoul of the Banach–Tarski paradox, you need a measure invariant not just under translations, but under arbitrary congruences (which is why @YCor's answer isn't ruled out). – LSpice Mar 28 '19 at 2:15
• Under congruences by the group of isometries or oriented isometries yields a "paradox" (no longer a paradox, but obstruction to existence of an invariant measure defined on all subsets) in dimension $\ge 3$. Using congruences by the group of determinant 1 or determinant $\pm 1$ (=measure-preserving) affine self-transformation yields such an obstruction in dimension $\ge 2$. – YCor Mar 28 '19 at 7:57

[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 $$t\in\mathbf{R}^*$$.

• 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? ($m(\mathbb{R}) = \infty$). – pitariver Feb 23 '19 at 19:20
• @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... – YCor Feb 23 '19 at 19:27
• 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. – pitariver Feb 23 '19 at 19:35
• I rewrote the answer using another approach. – YCor Feb 23 '19 at 19:43
• I think you got it, thanks again for the attention! – pitariver Feb 23 '19 at 20:05