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Unformatted text preview: that the sets that satisfy this condition form a σalgebra, so E is measurable if and only if E c is measurable. Since obviously * E ∈ S if and only if E c ∈ S , it suffices to prove the measurability of sets E such that E ⊂ (∞ , 0]. Assume thus E ⊂ (∞ , 0]. For every subset A of R , we’ll have sup( A ∩ E ) ≤ 0, thus μ * ( A ∩ E ) = 0. Assume first that A contains some x , x > 0. Then it is a simple exercise to prove that sup A = sup( A ∩ (0 , ∞ )) > 0, thus μ * ( A ) = sup( A ∩ (0 , ∞ )). We also have A ∩ (0 , ∞ ) = A ∩ E c ∩ (0 , ∞ ) and < sup( A ∩ (0 , ∞ )) = sup( A ∩ E c ∩ (0 , ∞ )) = sup( A ∩ E c ) = μ * ( A ∩ E c ); thus μ * ( A ) = μ * ( A ∩ E c ) if a ∩ (0 , ∞ ) 6 = ∅ . If A ∩ (0 , ∞ ) = ∅ , then A ⊂ (∞ , 0] and μ * ( A ) = 0, μ * ( A ∩ E c ) = 0 ≤ μ * ( A ) = 0 , so in all cases we have μ * ( A ) = μ * ( A ∩ E c ) Thus μ * ( A ) = 0 + μ * ( A ) = μ * ( A ∩ E ) + μ * ( A ∩ E c ) proving E is measurable. To conclude, we have to see that if E / ∈ S , then E is not measurable. Assume E / ∈ S . Then E 6⊂ (∞ , 0], thus there is x ∈ E , x > 0. But E 6 = F ∪ (0 , ∞ ) for a set F ⊂ (∞ , 0]; i.e., E 6⊃ (0 , ∞ ), hence there is y > 0, y / ∈ E . Let A = { x,y } . Then A ∩ E = { x } , hence μ * ( A ∩ E ) = sup( A ∩ E ) = x , μ * ( A ∩ E c ) = sup( A ∩ E ) = y , and μ * ( A ∩ E ) + μ * ( A ∩ E c ) = x + y > max( x,y ) = μ * ( A ). Thus E / ∈ S . (c) μ * ( A ) = 1 if A is uncountable, 0 if A is countable (or finite). Solution. The empty set is finite thus μ * ( ∅ ) = 0; OM1 holds. If A ⊂ B , if B is uncountable, then μ * ( A ) ≤ 1 = μ * ( B ); if B is countable, then so is A and μ * ( A ) = μ * ( B ) = 0. We see OM2 holds. If A n ⊂ R for all n ∈ N , if A n is uncountable for some n ∈ N , then ∑ ∞ n =1 μ * ( A n ) ≥ μ * ( A n ) = 1 ≥ μ * ( S ∞ n =1 A n ). On the other hand, if all A n ’s are countable, so is their union (a countable union of countable sets is countable) and ∑ ∞ n =1 μ * ( A n ) ≥ μ * ( A n ) = 0 = μ * ( S ∞ n =1 A n ). Let S consist of all sets that are either countable or that have countable complement. We claim this is precisely the σalgebra of measurable sets. Let E ∈ S . Suppose A ⊂ R . Assume first A is countable. Then both A ∩ E and A ∩ E c are countable and μ * ( A ) = 0 = μ * ( A ∩ E ) + μ * ( A ∩ E c ) . On the other hand if A is uncountable then precisely one of A ∩ E,A ∩ E c is uncountable, the other one countable. In fact, one has to be uncountable, because A is uncountable. But one has to be countable, because one of E,E c is countable. Thus μ * ( A ) = 1 = 1 + 0 = μ * ( A ∩ E ) + μ * ( A ∩ E c ) . This proves that all sets of S are measurable. Assume now that E / ∈ S . Since R is uncountable, the only way this is possible is if both E and its complement are uncountable. Then 1 = μ * ( R ) < 1 + 1 = μ * ( R ∩ E ) + μ * ( R ∩ E c ) , showing that E is not measurable....
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 Spring '11
 Speinklo
 Sets, Empty set, measure, Basic concepts in set theory, Lebesgue measure

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