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. 



Finitely Additive Homogeneous Translation Invariant Measure on $\mathcal{P}(\mathbb{R})$
[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 $t\in\mathbf{R}^\*$.


