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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 it's 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 which 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.

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.

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 it's 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 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 measure even exist? Any reference given is appreciated.

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

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 it's 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 which 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.