ia2sp10h7s

# 3 exercises from royden most are very easy i think

• Notes
• 4
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 2 - 4 out of 4 pages.

3. Exercises from Royden. (Most are very easy, I think. But they are instructive.) * To belabor the obvious; otherwise E (0 , ), then E c ( -∞ , 0], then E c is measurable, hence so is E . 2

Subscribe to view the full document.

(a) Chapter 3, # 4 (p. 55) (This is a very easy exercise, don’t go overboard with it. You may use the obvious, namely that if you have two disjoint sets A, B one with n elements, the other one with m elements, then A B has n + m elements. You may also use that if you have a sequence { A n } of pairwise disjoint sets, then the union is a finite set if and only if all sets are finite and all but a finite number of the sets A n are empty. The main point of assigning this exercise is so we can talk of the counting measure later on. (This exercise wasn’t graded. Everything is quite basic and I’ll skip a proof. (b) Chapter 3, # 5 (p. 58). Solution. Assume I 1 , . . . , I n are open intervals and Q [0 , 1] S n k =1 I k . We may assume all these in- tervals as being bounded;otherwise the conclusion holds trivially. Let I k = ( a k , b k ) for k = 1 , . . . , n . Let F = { a 1 , a 2 , . . . , a n , b 1 , b 2 , . . . , b n } be the set of all the endpoints of these intervals. Then F is a finite set, thus measurable and of measure 0. We notice that [0 , 1] \ F S n k =1 I k . In fact, let x [0 , 1] \ F . Since the rational numbers are dense in [0 , 1], there is a sequence { r m } in [0 , 1] Q converging to x . There being only a finite number of intervals I k , one of these intervals must contain an infinite number of terms of the sequence; i.e., a subsequence. But a subsequence in an interval ( a k , b k ) can only converge to an interior point or to an endpoint. Since x / F , we conclude it is an interior point in one of the intervals. Now 1 = m ([0 , 1]) = ([0 , 1] \ F ) + m ([0 , 1] F ) = ([0 , 1] \ F ) + 0 m n [ k =1 I k ! n X k =1 m ( I k ) = n X k =1 ( I k ) . (c) Chapter 3, # 7 (p. 58). Translating an open interval results in an open interval of exactly the same length. Solution. If I is an interval (open or not), it is a triviality to verify that I + x is an interval and ( I + x ) = ( I ) for all x R . It is also trivial to verify that if I n is an interval (or, actually any subset of R , and the union doesn’t even have to be countable), then ( bigcup n N I n ) + x = ( bigcup n N I n + x )) for all x R . But, trivial or not, these facts should be at the very least stated. Assume now A R . Let { I n } be a countable collection of open intervals such that A S n =1 I n . Then { I n + x } is a collection of intervals covering A + x and it follows that m * ( A + x ) X n =1 ( I n + x ) = X n =1 ( I n ) . Because { I n } was an arbitrary countable collection of open intervals such that A S n =1 I n , we proved that m * ( A + x ) is a lower bound of the set of (extended) numbers of which m * ( A ) is the infimum, thus m * ( A + x ) m * ( A ). Since A R , x R were arbitrary, we also have m * ( A ) = m * (( A + x ) + ( - x )) m * ( A + x ) for all A R , x R . Equality follows.
You've reached the end of this preview.
• Spring '11
• Speinklo
• Sets, Empty set, measure, Basic concepts in set theory, Lebesgue measure

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern