real-analysis-hw1-with-solution-2007-9-11 - Solutions to...

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Solutions to Homework 1 1. (10 points) Let Q be the set of all rationals in the interval [0 , 1] . Let S = { I 1 , I 2 , ..., I m } be a finite collection of closed intervals covering Q. Show that m X k =1 v ( I k ) 1 . (0.1) On the other hand, for any ε > 0 , one can fi nd S = { I 1 , I 2 , ..., I m , · · ·} , which is a count- able collection of closed intervals covering Q, such that X k =1 v ( I k ) < ε. (0.2) In particular, (0.2) implies that | Q | e = 0 . (Now you see the di ff erence between the use of ”finite cover” and ”countable cover” .) Solution: For (0.1), we fi rst assume that the intervals I 1 , I 2 , ... , I m are nonoverlapping . In such case we clearly have (0.1). For arbitrary intervals I 1 , I 2 , ... , I m with overlapping, one can throw away the over- lapping part and the remaining nonoverlapping part, which we denote it as J 1 , J 2 , ... , J n , satis fi es P n k =1 v ( J k ) 1 . Therefore we have (0.1). For (0.2), it has been done in class. ¤ 2. (10 points) Find a set E R with outer measure zero and a function f : E R such that f is continuous on E and f ( E ) = [0 , 1] . This exercise says that a
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  • Fall '08
  • Akhmedov,A
  • Continuous function, Lebesgue measure, Borel, Cantor Lebesgue, Cantor Lebesgue function

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