# Showing Lebesgue Measure is the unique measure on $\mathbb{R}^n$ such that it is translation invariant

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.

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

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 $$\mathcal{K}$$.

• By almost disjoint I mean that the interior of the cubes don't intersect, in other words, the cubes intersect only along faces. Oct 20, 2020 at 5:08
• The fact that $\mu$ is finite on compact subsets follows from the unit cube having finite measure. Oct 20, 2020 at 5:23
• But why should the fact that $\mu$ is finite on compact subsets imply that it is regular? Oct 20, 2020 at 5:24
• @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. Oct 20, 2020 at 5:26