0
$\begingroup$

My approach is to assume we have some other measure and prove that it must be the Lebesgue measure, so assume such a measure exists. Unfortunately from here I am quite stuck, any tips would be much appreciated.

$\endgroup$
1
  • $\begingroup$ This not a simple result. It is proved in books on Haar measure. You should not even attempt to prove ii from the definition. $\endgroup$ Oct 20, 2020 at 5:07

1 Answer 1

6
$\begingroup$

Lebesgue measure is the unique (Borel) measure which is translation invariant, finite on compact sets and attains 1 on the unit cube (without this last one, you can scale the Lebesgue measure by some $c>0$ and still have a translation invariant measure). Here is a sketch of the proof.

As far as I know, you need finiteness on compact sets, so that the measure is regular. Once you have that, you just need to know the measure on open sets. But in $\mathbb{R}^n$ open sets are almost disjoint union of cubes, so you only need the measure on cubes.

Let $\mu$ be a translation invariant Borel measure finite on compact sets and let $c = \mu([0, 1]^n)$. Set $\mathcal{K} = \{E\in \mathcal{B}:\mu(E) = c\lambda(E)\}$ where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathcal{R}^n$ and $\lambda$ is the Lebesgue measure.

Divisibility property: Under translational invariance if a Borel measurable set $E = \sqcup_{i=1}^m (t_i+F)$ is written as a disjoint union of finite translates of another Borel measurable $F$, then you can show that $E\in\mathcal{K}\implies F\in\mathcal{K}$.

You should also show that $\mu$ is zero on subsets of hyperplanes (obtained by setting some coordinate = constant). This can be shown by first translating the plane so that it becomes $x_i = 0$ say, then this plane is the countable union of translates of $Q' = (0, 1]^{n-1}\times\{0\}$ where I have taken $i = n$ for convenience. So it suffices to show $\mu(Q') = 0$.

This follows because (disjoint) countable translates of $Q'$ (for example, $\sqcup_{n\geq 1}(1/n+Q')$) are contained in $Q$ and $Q$ has finite $\mu$ measure, so $\mu(Q') = 0$.

Given an open set, write it as an almost disjoint union of cubes with rational side lengths (in fact with side length $= 1/2^k, k\geq 1$). Because hyperplanes have $\mu$ measure $0$, you can treat them as disjoint union of cubes. By the divisibility property all rational side length cubes are in $\mathcal{K}$, therefore all open sets are also in $\mathcal{K}$.

$\endgroup$
4
  • $\begingroup$ By almost disjoint I mean that the interior of the cubes don't intersect, in other words, the cubes intersect only along faces. $\endgroup$ Oct 20, 2020 at 5:08
  • $\begingroup$ The fact that $\mu$ is finite on compact subsets follows from the unit cube having finite measure. $\endgroup$ Oct 20, 2020 at 5:23
  • $\begingroup$ But why should the fact that $\mu$ is finite on compact subsets imply that it is regular? $\endgroup$ Oct 20, 2020 at 5:24
  • 1
    $\begingroup$ @AndréPorto yes, they are equivalent. I assumed finiteness on compact sets to have regularity. Regularity follows from the Riesz-Markov-Kakutani representation theorem. Basically finiteness on compact sets gives you a linear nonnegative functional $C_C(\mathbb{R}^n)\to\mathbb{C}$ where the left side is all compactly supported continuous functions. $\endgroup$ Oct 20, 2020 at 5:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .