Lebesgue Measurable Sets

Transcription

1 s Dr. Aditya Kaushik Directorate of Distance Education Kurukshetra University, Kurukshetra Haryana India

2 s

3 Definition A set E R is said to be Lebesgue measurable if for any A R we have m A) = m A ) E + m A E c). 1) We may write the above equality as a combination of following two inequalities 1 m A) m A E) + m A E c ). 2 m A) m A E) + m A E c ). Is there any wild guess!!! Why we are doing this?

4 Write, A = A E) A E) c, we then have [ m G) = m A ) E A ) c ] E m A ) E + m A c E). 2) Are you able conclude anything from here? I think yes!!! If NO is your answer, I suggest you to look back to equality 1). Then, have a look at 2). What do you think is left to prove 1) if 2) holds eventually? A necessary and sufficient condition for E to be measurable is that for any set A R m A) m A ) E + m A E c).

5 Let us begin with the very simple yet important lemma: Lemma If m E) = 0, then E is measurable. Proof. Let A be any set of real numbers. Then, A E E m A E) m E) and A E c A m A E c ) m A). Therefore, m A E Hence, E is measurable. ) + m A E c) m E) + m A), = 0 + m A).

6 Suppose we are given with a measurable set, then what about its complement? Lemma E is measurable iff E c is measurable. Proof. Let A R and E be a measurable set. Then, m A) = m A ) E + m A E c) = m A E c) + m A E cc), [ E = E cc ]. Therefore, E c is measurable.

7 Proof Continues. Conversely, suppose that E c is measurable. Then m A) = m A E c) + m A E cc) = m A E c) + m A E cc) [ E c c = E ]. Hence, E is measurable.

8 Moreover, in the next theorem we see that union of two measurable sets is again a measurable set. Theorem Let E 1 and E 2 be two measurable sets, then E 1 E2 is measurable. Proof. Let A be any set of reals and E 1, E 2 be two measurable sets. Since E 2 is m able we have m A ) E1 c = m A ) E1 c E2 + m A ) E1 c E c 2.

9 Proof Coninues. Now, A E 1 E2 ) = = [A E 1 ] [A E 2 ] [A E 1 ] [A E 2 E c 1 ]. m A E 1 E2 )) m A E 1 ) + m [ A E 2 E c 1 ]

10 Proof Continues. Let us now consider m A E 1 E2 )) + m A E 1 E2 ) c ) m A E 1 ) + m A E c 1 = m A). Since E 1 is also given measurable. Hence, E 1 E2 is also measurable. )

11 Definition A class a of sets is said to be an algebra if it satisfies the following conditions: 1 If E a then E c a. 2 If E 1 and E 2 a, then E 1 E2 a. Thus a class a of sets is said to be algebra if it is closed under the formation of complements or finite unions.

12 Lemma Algebra is closed under the formation of finite intersections. Proof. Let A 1, A 2,...,A n a. Then, A 1 A2... A n ) c = A c 1 A c 2... A c n. Now, A n a n, then A c n a because a is algebra and is therefore closed under the formation of compliments.

13 Proof Continues. Further, a being algebra is closed under the formation finite unions. This implies that A c 1 A c 2... A c n a { A 1 A2... An } c a. It follows that A 1 A2... An a.

14 Definition σ-algebra) A class a is said to be σ-algebra, if it is closed under the formation of countable unions and of complements. It is an easy exercise for the readers to verify that that σ-algebra is closed under the formation of finite intersection.

15 Theorem Let A be any set of real number and let E 1, E 2,...,E n be pair-wise disjoint Lebesgue measurable sets then m A )) n E i = m A ) E i.

16 Proof. We prove the result using mathematical induction on n. For, n = 1 m A ) E 1 = m A ) E 1 Thus, the result is true for n = 1. Suppose that the result is true for n-1) sets E i then we have n m A n ) = m A n Ei j=1 E n j=1

17 Proof Continues. Now, since E i s are disjoint, we have A E n = A E n. j=1 E n And A [ n ] En c E i == A n 1 E i ).

18 Proof Continues. It follows that m and A [ n ] ) E i En = m A E n ), m A [ n ] ) E i E c n = m A [ n 1 ]) E n E i. Addition of above two equations leads us to the required results.

19 Theorem Countable union of measurable sets is measurable. Proof. Let {A n } be any countable condition of measurable sets and E = n=1 A n. We know that the class of Lebesgue measurable set constitutes algebra. Therefore, there is a sequence {E n } of pair-wise disjoint measurable sets such that E = A n = E n. n=1 n=1 Let F n = E i, then F n is measurable for each n and F n E. This implies thatf c n E c.

20 Proof Continues. Moreover, if A be any set of real numbers then A ) Fn c A E c) m A E c) m A F c n ).

21 Proof Continues. Since, F n is measurable we have m A) m A ) F n + m A ) Fn c, m A [ n ]) E i + m A E c), = n m A ) E i + m A E c). L.H.S. being independent of n, it follows that m A) m A ) E i + m A E c) 3)

22 Proof Continues. Now, A [ ]) E i = Therefore m A [ ]) E i = m m A ) E A E i ). A E i ) ) m A ) E i m A ) E i 4)

23 Proof Continues. Combining 3) and 4), it gives m A) m A E ) + m A E c). Hence,E = E i is measurable. As a consequence of result we just proved, we have Corollary The class of Lebesgue measurable sets is a σ algebra.

24 Let us end this lecture with the statement of an important results. Theorem Interval a, ) is measurable. Proof of the above theorem is left for the readers as an exercise.

25 G.de Barra : Measure theory and integration, New Age International Publishers. A. Kaushik, Lecture Notes, Directorate of Distance Education, Kurukshetra University, Kurukshetra.

26 Thank You!