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Unformatted text preview: Solution 5 1. • Prove that F is a σfield: ∅ is countable = ⇒ ∅ , Ω ∈ F ; A ∈ F = ⇒ A or A c is countable = ⇒ A c ∈ F ; If { A i } ∞ i =1 ⊂ F , then either each A i is countable or there exists some A j which is uncountable. In the first case, S A i is countable. In the second case, A c j is countable, so ( S A i ) c = T A c i ⊂ A c j is countable. Hence in both case, S A i ∈ F . • Prove that μ is a measure: μ ( ∅ ) = 0 since ∅ is countable; Suppose { A i } ∞ i =1 ⊂ F are disjoint. If each A i is countable, then S A i is countable, so μ ( S A i ) = 0 = ∑ i μ ( A i ). If there exists some A j is uncoutable, then A c j is countable, S i 6 = j A i ⊂ A c j is countable and hence for each i 6 = j , A i is countable. So μ ( [ A i ) = 1 = μ ( A j ) + μ ( [ i 6 = j A i ) = X μ ( A i ) • Classify the measurable functions and μintegrable functions: Claim: f : Ω→ B ( R ) is measuable ⇐⇒ f is constant on some uncountable set. Proof: ” ⇐ ” is obvious since if f = a on uncoutable set E , then { f > b } is countable if b ≥ a and { f > b } c = { f ≤ b } is countable if b < a . In both case, { f > b } ∈ F . Hence f is measurable. ” ⇒ ”: Note when f is measurable, there is atmost one point a ∈ F such that { f = a } is uncountable, because if a 6 = b , then { f = a } and { f = b } are disjoint sets in F and atmost one of them is uncoutable....
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 Spring '08
 AMJAD
 lim, fn dµ, Dominated convergence theorem, Fatou's lemma

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