# Translation-Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure

## Theorem

Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\mathcal B \left({\R^n}\right)$.

Suppose that $\mu$ is translation-invariant.

Also, suppose that $\kappa := \mu \left({\left[{0 \,.\,.\, 1}\right)^n }\right) < +\infty$.

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

## Proof

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

$\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{J}^n_{ho,\text{rat}}}\right)$

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

So let $J = \left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal{J}^n_{ho,\text{rat}}$.

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 $\left[{0 \,.\,.\, \dfrac 1 M}\right)^n$, in that:

$\displaystyle J = \bigcup_{i \mathop = 1}^{k \left({J}\right)} \mathbf x_i + \left[{0 \,.\,.\, \frac 1 M}\right)^n$

for some $k \left({J}\right) \in \N$ and suitable $\mathbf x_i$, where:

$\mathbf x_i + \left[{0 \,.\,.\, \dfrac 1 M}\right)^n := \left[\left[{\mathbf x_i \,.\,.\, \mathbf x_i + \dfrac 1 M}\right)\right)$

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

$\mu \left({J}\right) = k \left({J}\right) \, \mu \left({\left[{0 \,.\,.\, \dfrac 1 M}\right)^n}\right)$
$\lambda^n \left({J}\right) = k \left({J}\right) \, \lambda^n \left({\left[{0 \,.\,.\, \dfrac 1 M}\right)^n}\right)$

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

$k \left({I}\right) = M^n$

For brevity, write $I/M$ for $\left[{0 \,.\,.\, \dfrac 1 M}\right)^n$.

Now, using that:

$\displaystyle \lambda^n \left({I/M}\right) = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$

and $\mu \left({I}\right) = \kappa$, compute:

 $\displaystyle \mu \left({J}\right)$ $=$ $\displaystyle k \left({J}\right) \mu \left({I/M}\right)$ Definition of $k \left({J}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {k \left({J}\right)} {M^n} \left({M^n \mu \left({I/M}\right)}\right)$ $M^n / M^n = 1$ $\displaystyle$ $=$ $\displaystyle \frac {k \left({J}\right)} {M^n} \mu \left({I}\right)$ $M^n = k \left({I}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\kappa \, k \left({J}\right)} {M^n}$ Definition of $\kappa$ $\displaystyle$ $=$ $\displaystyle \kappa \, k \left({J}\right) \, \lambda^n \left({I/M}\right)$ $\displaystyle$ $=$ $\displaystyle \kappa \, \lambda^n \left({J}\right)$ Definition of $k \left({J}\right)$

Therefore, $\mu \left({J}\right) = \kappa \, \lambda^n \left({J}\right)$ for all $J \in \mathcal{J}^n_{ho,\text{rat}}$.

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 $\left[{-k \,.\,.\, k}\right)^n \mathop \uparrow \R^n$

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

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

$\blacksquare$