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Unformatted text preview: Real Analysis - Math 630 Homework Set #1 by Bobby Rohde 9-7-00 Problem 1 & If A and B are two sets in & with A & B , then mA mB. & Proof Since & is a &-algebra, we know that C = B & A & , with C & A = and C A = B . Hence mB = m ( C A ) = mC + mA mA . mB mA, QED. Problem 2 & Let < E n > be any sequence of sets in & . Then m ( & E n ) mE n . & Proof By Proposition 1.2, a sequence < A n > of sets in & with E n = A n and A n & A m = , n m. Thus m ( E n ) = m ( A n ) = mA n . If we construct A n from E n according to the algorithim used in the proof of Proposition 1.2 then we have A n E n n, and from Problem 1 therefore mA n mE n . Hence we have mA n mE n m ( E n ) mE n . QED. Problem 3 & If there is a set A in & such that mA < & , then m = 0. & Proof By Way of Contradiction (BWOC). Suppose m & = 0, and let mA = < . A & & = & , hence A and & are disjoint. Thus by Property 3, m ( A & ) = mA + m & = + . But A & = A , so m ( A & ) = mA = . Thus = + > . Which is a contradiction, hence m & = 0. QED. Problem 4 & Let nE be & for an infinite set E and equal to the number of elements in E for a finite set. Show that n is a countably additive set function that is translation invariant and defined for all sets of real numbers. translation invariant and defined for all sets of real numbers....
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