Is Lebesgue measure the only $\sigma$-finite invariant Borel measure?

TL;DR

This paper investigates whether Lebesgue measure is uniquely characterized as the only translation-invariant σ-finite Borel measure on ℝ^d, revealing that the answer varies based on measure definitions and extensions.

Contribution

It clarifies the conditions under which Lebesgue measure is unique, showing that the answer depends on the measure's domain and the interpretation of the constant.

Findings

If defined only on Borel sets, Lebesgue measure is unique.

Extending the measure to a larger σ-algebra can produce non-unique measures.

Allowing the constant to be infinity makes Lebesgue measure effectively unique.

Abstract

R.D.Mauldin asked if every translation invariant $\sigma$-finite Borel measure on $\RR^d$ is a constant multiple of Lebesgue measure. The aim of this paper is to show that the answer is "yes and no", since surprisingly the answer depends on what we mean by Borel measure and by constant. We present Mauldin's proof of what he called a folklore result, stating that if the measure is only defined for Borel sets then the answer is affirmative. Then we show that if the measure is defined on a $\sigma$-algebra \emph{containing} the Borel sets then the answer is negative. However, if we allow the multiplicative constant to be infinity, then the answer is affirmative in this case as well. Moreover, our construction also shows that an isometry invariant $\sigma$-finite Borel measure (in the wider sense) on $\RR^d$ can be non-$\sigma$-finite when we restrict it to the Borel sets.