Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure

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Theorem

Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.

Suppose that $\mu$ is translation invariant.

Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\infty$.


Then $\mu = \kappa \lambda^n$, where $\lambda^n$ is the $n$-dimensional Lebesgue measure.


Proof

From Characterization of Euclidean Borel Sigma-Algebra, we have:

$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$

where $\JJ^n_{ho, \text {rat} }$ denotes the collection of half-open $n$-rectangles with rational endpoints.


So let $J = \horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho, \text {rat} }$.

Let $M \in \N$ be a common denominator of the $a_i, b_i$ (which are rational by assumption).

We may then cover $J$ by finitely many pairwise disjoint half-open $n$-rectangles $\hointr 0 {\dfrac 1 M}^n$, in that:

$\ds J = \bigcup_{i \mathop = 1}^{\map k J} \mathbf x_i + \hointr 0 {\dfrac 1 M}^n$

for some $\map k J \in \N$ and suitable $\mathbf x_i$, where:

$\mathbf x_i + \hointr 0 {\dfrac 1 M}^n := \horectr {\mathbf x_i} {\mathbf x_i + \dfrac 1 M}$


Using that $\mu$ is translation invariant, this means:

$\map \mu J = \map k J \, \map \mu {\hointr 0 {\dfrac 1 M}^n}$

Also, by Lebesgue Measure is Translation Invariant:

$\map {\lambda^n} J = \map k J \, \map {\lambda^n} {\hointr 0 {\dfrac 1 M}^n}$

A moment's thought shows us that the half-open $n$-rectangle $I = \hointr 0 1^n$ may be covered by $M^n$ copies of $\hointr 0 {\dfrac 1 M}^n$, so that:

$\map k I = M^n$

For brevity, write $I / M$ for $\hointr 0 {\dfrac 1 M}^n$.

Now, using that:

$\ds \map {\lambda^n} {I / M} = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$

and $\map \mu I = \kappa$, compute:

\(\ds \map \mu J\) \(=\) \(\ds \map k J \map \mu {I / M}\) Definition of $\map k J$
\(\ds \) \(=\) \(\ds \frac {\map k J} {M^n} \paren {M^n \map \mu {I / M} }\) $M^n / M^n = 1$
\(\ds \) \(=\) \(\ds \frac {\map k J} {M^n} \map \mu I\) $M^n = \map k I$
\(\ds \) \(=\) \(\ds \frac {\kappa \, \map k J} {M^n}\) Definition of $\kappa$
\(\ds \) \(=\) \(\ds \kappa \, \map k J \, \map {\lambda^n} {I / M}\)
\(\ds \) \(=\) \(\ds \kappa \, \map {\lambda^n} J\) Definition of $\map k J$

Therefore, $\map \mu J = \kappa \, \map {\lambda^n} J$ for all $J \in \JJ^n_{ho, \text {rat} }$.


Let us quickly verify the other conditions for Uniqueness of Measures.

For $(1)$, we have Half-Open Rectangles Closed under Intersection.

For $(2)$, observe the exhausting sequence $\hointr {-k} k^n \mathop \uparrow \R^n$

Finally, for $(4)$, we recall that $\kappa < +\infty$.


Thus, by Uniqueness of Measures, $\mu = \kappa \lambda^n$.

$\blacksquare$


Sources

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