TheoremBase

Existence and Uniqueness of Lebesgue Measure on ℝⁿ

Assuming dependent choice, the open boxes of ℝⁿ form a π-system generating the Borel sets, two measures agreeing and finite on open boxes are equal, and exactly one Borel measure on ℝⁿ gives every nonempty open box its volume.

Statement

In the setting of Measurable Functions and the Lebesgue Integral: Notation, assume the axiom of dependent choice. Let n∈Nn\in\mathbb{N}, let R(a,b)R(a,b), for a,b∈Rna,b\in\mathbb{R}^{n}, be the open boxes of Rn\mathbb{R}^{n}, and let Rn\mathcal{R}_{n} be the set of open boxes.

Boxes generate the Borel sets. Rn\mathcal{R}_{n} is a π\pi-system on Rn\mathbb{R}^{n} with

σ(Rn)=B(Rn).\sigma(\mathcal{R}_{n})=\mathcal{B}(\mathbb{R}^{n}).

Cubes cover Rn\mathbb{R}^{n}. For k∈Nk\in\mathbb{N} let Qk=R(−ku,ku)Q_{k}=R(-k\mathbf{u},k\mathbf{u}), where u=(1,…,1)∈Rn\mathbf{u}=(1,\dots,1)\in\mathbb{R}^{n} is the point with every component 11. Then

Rn=⋃k∈NQk.\mathbb{R}^{n}=\bigcup_{k\in\mathbb{N}}Q_{k}.

Existence and uniqueness. There is exactly one measure κ\kappa on (Rn,B(Rn))(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n})) such that

κ(R(a,b))=∏i=1n(bi−ai)\kappa\bigl(R(a,b)\bigr)=\prod_{i=1}^{n}(b_{i}-a_{i})

for all a,b∈Rna,b\in\mathbb{R}^{n} with ai<bia_{i}<b_{i} for every i∈[n]i\in[n].

Boxes determine measures. If μ\mu and ν\nu are measures on (Rn,B(Rn))(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n})) with μ(R)=ν(R)<∞\mu(R)=\nu(R)<\infty for every R∈RnR\in\mathcal{R}_{n}, then μ=ν\mu=\nu.

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