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In L. Simon's book "Introduction to Geometric Measure Theory", one can find the following characterization of the Lebesgue outer measure (in $\mathbb{R}^n$) on page 11:

It is the unique outer measure $\mathcal{L}^n$ such that

  1. $\mathcal{L}^n\left( \prod_{i=1}^n (a_i, b_i) \right) = \prod_{i=1}^n (b_i - a_i)$ for all open boxes;
  2. for all subset $A \subset \mathbb{R}^n$, we have $\mathcal{L}^n(A) = \inf \left\{ \mathcal{L}^n(U) : A \subset U \text{ and } U \text{ open} \right\}$.

I have trouble proving this statement. The difficulty is that this characterization makes no explicit assumption that $\mathcal{L}^n$ be a Borel regular outer measure. Instead, one needs to prove that open sets are measurable (in the sense of Carathéodory), as a consequence of properties 1 and 2.

vizietto
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1 Answers1

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Let $L$ be Lebesgue measure in $\mathbb R^n$, and let $m$ be another outer measure that satisfies properties (1) and (2).

Step 1. For any product of half-open intervals $B_1=\prod_{i=1}^n [a_i,b_i)$, we have $ \prod_{i=1}^n (a_i ,b_i) \subset B_1 \subset \prod_{i=1}^n (a_i-\epsilon ,b_i) $ for all $\epsilon>0$, so $m(B_1)=L(B_1)$ by monotonicity.

Step 2. Let $U$ be a bounded open set, so that $U \subset Q$ for some half-open dyadic cube $Q$. Let $E$ be a finite union of half-open dyadic cubes contained in $U$. Then $m(E) \le L(E)$ by subadditivity, and similarly, $m(Q\setminus E) \le L(Q\setminus E)$, as $Q\setminus E$ is also a finite union of half-open dyadic cubes. Since $$L(Q)=m(Q) \le m(E)+m(Q\setminus E) \le L(E)+L(Q\setminus E) =L(Q) \,,$$ we deduce that $m(E)=L(E)$.

Step 3. Let $U,Q$ be as in step 2. Let $E_k=E_k(U)$ be the union of all half-open dyadic cubes of side $1/2^k$ contained in $U$. Then $$U=E_0 \cup \Bigl(\bigcup_{k=1}^\infty (E_k \setminus E_{k-1})\Bigr) \,.$$ Subadditivity implies that $$m(U) \le m(E_0)+ \sum_{k=1}^\infty m(E_k \setminus E_{k-1}) \,,$$ so (using this for the upper bound and monotonicity for the lower bound) $$m(U)=\lim_k m(E_k)=\lim_k L(E_k)=L(U)\,.$$

Step 4. If $A \subset \mathbb R^n$ is bounded, then step 3 and property 2 imply that $m(A)=L(A)$.

Step 5. Given any $A \subset \mathbb R^n$, write $A_j=A \cap [-j,j]^n$. Then $$m(A) \le m(A_1)+ \sum_{j=2 }^\infty m(A_j \setminus A_{j-1}) \,,$$ so (using this for the upper bound and monotonicity for the lower bound) $$m(A) =\lim_j m(A_j)=\lim_j L(A_j)=L(A) \,.$$

Yuval Peres
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