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Unformatted text preview: Analysis Midterm 1 Solutions 1. Let (Ω , F , μ ) be a measure space on which f, f 1 , f 2 , . . . are nonnegative integrable functions. Show that if f n → f almost everywhere and R f n dμ → R fdμ then R  f n f  dμ → 0. Solution Notice that 0 6 ( f n f ) 6 f since f, f n are nonnegative. The Dominated Convergence Theorem therefore implies that R ( f n f ) dμ → since f n → f μae and the dominating function f is integrable. Now observe that Z  f n f  dμ = Z ( f n f ) dμ + 2 Z ( f n f ) dμ tends to 0 since on the right the first summand does so (given) and the second summand does so (shown). [This result is often called Sheff´ e’s lemma .] 2. Let (Ω , F , μ ) be a probability space. For Z ⊂ Ω define μ * ( Z ) = sup { μ ( A ) : F 3 A ⊂ Z } μ * ( Z ) = inf { μ ( B ) : Z ⊂ B ∈ F} define G = { Z ⊂ Ω : μ * ( Z ) = μ * ( Z ) } and for Z ∈ G write ν ( Z ) = μ * ( Z ) = μ * ( Z ). Prove that: (i) Z ∈ G if and only if for each ε > 0 there exist A, B ∈ F such that A ⊂ Z ⊂ B and μ ( B A ) < ε ; (ii) G is a σalgebra (containing F ) on which ν is a measure (extending μ ). Solution (i) Let ε > 0. ( ⇒ ) If Z ∈ G then there exist F 3 A ⊂ Z such that ν ( Z ) ε/ 2 < μ ( A ) and Z ⊂ B ∈ F such that μ ( B ) < ν ( Z ) + ε/ 2; now μ ( B A ) = μ ( B ) μ ( A ) < ε . ( ⇐ ) Let A, B ∈ F satisfy A ⊂ Z ⊂ B and μ ( B A ) < ε : then μ * ( Z ) > μ ( A ) > μ ( B ) ε > μ * ( Z ) ε so that μ * ( Z...
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 Fall '09
 Robinson
 µ, Hilbert space, normed vector space, Dominated convergence theorem

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