HWsolution1

HWsolution1 - Math 447 Solutions to Assignment 1 Spring...

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Math 447 Solutions to Assignment 1 Spring 2009 Exercise 16.4 on page 271 Given any subset of E of R and any h R , show that m * ( E + h ) = m * ( E ), where E + h = { x + h : x E } . Solution The statement, which is part (iii) of Proposition 16.2 on page 269, says that Lebesgue outer measure is translation invariant. Since outer measure is defined in terms of coverings by intervals, and since translation preserves the lengths of intervals, the statement is supposed to be intuitively clear. Here is a formal argument. If E is covered by a sequence of intervals I n , then the intervals I n + h have the same total length and cover the set E + h . Hence m * ( E + h ) m * ( E ). Conversely, if a sequence of intervals J n covers the set E + h , then the intervals J n - h have the same total length and cover the set E . Hence m * ( E ) m * ( E + h ). Putting the two inequalities together shows that m * ( E ) = m * ( E + h ). Exercise 16.8 on page 271 Given δ > 0, show that m * ( E ) = inf n =1 ( I n ) where the infimum is taken over all coverings of E by sequences of intervals ( I n ), where each I n has diameter less than δ . Solution Notice that the indicated property differs from the definition of outer measure at the top of page 269 only in the requirement that the intervals I n have uniformly small length. Therefore it suffices to show that every countable collection of intervals ( I n ) can be replaced by another countable collection of intervals ( J n ) such that S n I n = S n J n and n ( I n ) = n ( J n ) and ( J n ) < δ for every n . Since a countable collection of finite sets is countable (a special case of Theorem 2.6 on page 21), it suffices to show that a single interval can be partitioned into a finite union of subintervals each of length less than δ . That such a partition exists should be obvious, but here is a formal argument. One can construct the partition concretely through repeated bisection (for example). If the interval has length , then dividing the interval at the midpoint gives 2 subintervals each of length ‘/ 2; bisecting again gives 2 2 subintervals each of length ‘/ 2 2 ; and so on. After k steps there are 2 k subintervals each of length ‘/ 2 k . The construction can be stopped as soon as ‘/ 2 k < δ , that is, k > log 2 ( ‘/δ ). Notice that since the length of an interval, the length of its closure, and the length of its interior are all equal, it is not necessary to worry in the preceding argument about which subintervals contain the partition points. Exercise 16.9 on page 271
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HWsolution1 - Math 447 Solutions to Assignment 1 Spring...

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