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NnbljG-13

# NnbljG-13 - MAT 473 33 13 Lebesgue measure inner and outer...

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MAT 473 33 13. Lebesgue measure: inner and outer measures Proposition 13.1. Let A i be subsets of R N . (i) If A 1 A 2 , then λ ( A 1 ) λ ( A 2 ) and λ ( A 1 ) λ ( A 2 ) . (ii) λ i =1 A i i =1 λ ( A i ) . (iii) If the A i are disjoint, then λ i =1 A i i =1 λ ( A i ) . Proof. (i) is routine. For (ii), let > 0 and for each i choose an open set G i containing A i with λ ( G i ) < λ ( A i ) + / 2 i . Then i G i is an open set containing i A i , so λ i =1 A i λ i =1 G i i =1 λ ( G i ) < i =1 ( λ ( A i ) + / 2 i ) = i =1 λ ( A i ) + . Since > 0 was arbitrary, we have λ ( i A i ) i λ ( A i ). Similarly, for (iii), fix N N and for each i let K i be a compact subset of A i . Then N i K i is a compact subset of i A i , and since the union of K i ’s is finite and disjoint we have λ i =1 A i λ N i =1 K i = N i =1 λ ( K i ) . Taking the sup over all such K i gives λ ( i A i ) N i λ ( A i ), and now since N was arbitrary we have λ ( i A i ) i λ ( A i ). Definition 13.2. Let L 0 = { A R n | λ ( A ) = λ ( A ) < } . For A L 0 we set λ ( A ) = λ ( A ). By Exercise 12.4, all the compact sets and all the open sets with finite measure are in L 0 , and the definition of λ ( A

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