s Dr. Aditya Kaushik Directorate of Distance Education Kurukshetra University, Kurukshetra Haryana 136119 India
s
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?
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).
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).
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.
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.
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.
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 ]
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. )
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.
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.
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.
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.
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.
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
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 ).
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.
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.
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 ).
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)
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)
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.
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.
G.de Barra : Measure theory and integration, New Age International Publishers. A. Kaushik, Lecture Notes, Directorate of Distance Education, Kurukshetra University, Kurukshetra.
Thank You!