18.100pset10

18.100pset10 - 18.100B/C: Fall 2008 Solutions to Homework...

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 10 1. (a) Suppose that N R such that N C is not dense in R . Then there is some point x N that is not a limit point of N C . Thus, there is an > 0 such that B ( x ) N . It follows that any collection of balls that cover N must cover B ( x ) and so must have radii that sum to at least , and so N cannot have measure 0. Taking the contrapositive gives our result. (b) Suppose that N R is open and nonempty. Then we may choose x N and > 0 so that B ( x ) N . Then the rest of the proof is as in part (a). (c) Yes: for example, N R is closed and noncompact but is measure 0: to cover N with open intervals of total radius less than , take B n = B / 2 n +1 ( n ) for n N . Similarly we take Q as an example of a dense set of measure 0: let : N Q be a bijection, and take B n = B / 2 n +1 ( ( n )) for n N . (This also shows that every countable subset of R is measure 0.) 2. First, note that g is not bounded and so cannot be Riemann integrable. Now we turn our attention to f . For any partition P of [0 , 1], L ( P,f ) = 0. By Rudins 6.6, it suffices to show that for each > 0, there is a partition for which U ( P,f ) < . Choose so that < 2 12 , and choose a partition 0 = x < x 1 < ... < x k < x k +1 < ... < x m = 1 where x i +1 x i < and such that k is the smallest integer such that x k > 2 . Then the only intervals of our partition that contain a point 1 n for some n are the intervals [ x i ,x i +1 ] for 0 i k 1 and the intervals that contain 1 n for 1 n 2 . Thus U ( P,f ) x k + 2 2 < ( 2 + ) + 3 < , as claimed. Thus inf U ( P,f ) = 0 = sup L ( P,f ) and so integraltext 1 f ( x ) dx = 0....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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18.100pset10 - 18.100B/C: Fall 2008 Solutions to Homework...

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