smt1 - Math 8601. October 6, 2010. Midterm Exam 1. Problems...

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Math 8601. October 6, 2010. Midterm Exam 1. Problems and Solutions. Problem 1. Let ( X, M ) be a measure space with μ ( X ) < . Show that for arbitrary A,B,C ∈ M , we have | μ ( A B ) - μ ( A C ) | ≤ μ ( B Δ C ) . where B Δ C := ( B \ C ) ( C \ B ) = ( B C ) \ ( B C ) - the symmetric difference of B and C . Proof. One can use the triangle inequality for the “distance” ρ ( A,B ) := μ ( A Δ B ), This property simply means that ρ ( A,B ) := μ ( A Δ B ) satisfies the triangle inequality, which is Problem #5 in HW #2. It is easy to see that B 0 := A B and C 0 := A C satisfy ( B 0 Δ C 0 ) = A ( B Δ C ) ( B Δ C ). Then | μ ( A B ) - μ ( A C ) | = | μ ( B 0 ) - μ ( C 0 ) | = | ρ ( B 0 , ) - ρ ( C 0 , ) | ≤ ρ ( B 0 ,C 0 ) = μ ( B 0 Δ C 0 ) μ ( B Δ C ) . Alternatively, one can note that μ ( A B ) = μ ( A B C ) + μ ( A B C c ) , μ ( A C ) = μ ( A B C ) + μ ( A B c C ) , hence | μ ( A B ) - μ ( A C ) | ≤ μ ( A B C c ) + μ ( A B c C ) μ ( B C c ) + μ ( B c C ) = μ ( B Δ C ) . Problem 2. For A, B R 1 , define the algebraic sum A + B = { x R 1 : x = a + b for some a A, b B } . Either prove or give a counterexample for each of the following statements.
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smt1 - Math 8601. October 6, 2010. Midterm Exam 1. Problems...

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