msreinteg

# msreinteg - Lecture 2. Measure and Integration There are...

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Unformatted text preview: Lecture 2. Measure and Integration There are several ways of presenting the definition of integration with respect to a measure. We will follow, more or less, the approach of Rudins Real and Complex Analysis - this is probably the fastest route. Conventions concerning . In measure theory it is often essential to use and- . We shall view these as two objects and define the following relationship with real numbers (the set of all real numbers will be denoted by R ) :- < c < for every c R c + = if- < c , and c + (- ) =- if- c < = 0 c ( ) = , if 0 < c ; , if- c < 0; Note that - is not defined! The convention 0 = 0 might seem like an arbitrary choice, but it is the really the right one : it will be useful in integration and there it will appear as the natural choice. These operations have the usual properties on [0 , ] : commutative, associative and distributive. Throughout our discussion ( X, B , ) will be a measure space. This means that X is a set, B is a collection of subsets of X forming a - algebra, and is a measure. Recall that to say that B is a - algebra means that : (S1) B ; (S2) if A B then the complement A c B ; (S3) if A 1 , A 2 , A 3 , ... is a countable collection of sets with each A i B then i =1 A i B . Sets in B will be called B- measurable sets, or simply measurable sets. Observations . (i) The intersection of a countable collection A 1 , A 2 , ... of measurable sets is also measurable (because i =1 A i = ( i A c i ) c is measurable by (S2) and (S3).) (ii) If A 1 , ..., A m B then m i =1 A i B ; this follows from (S1) and (S3) by taking A n = for all n bigger. Using the complements we obtain also that m i =1 A i B . (iii) Notice also that if A, B B then A \ B , being equal to A B c , is also in B . To say that is a measure on ( X, B ) means that is a mapping B [0 , ] such that : (M1) ( ) = 0 (M2) if A 1 , A 2 , ... is a countable collection of mutually disjoint sets in B (i.e. each A i B and A i A j = for every i 6 = j ) then : ( i =1 A i ) = X i =1 ( A i ) . Taking A i = for i bigger than some m , we see (from (M1) and (M2)) that is finitely-additive : 1 ( m i =1 A i ) = m X i =1 ( A i ) provided, of course, that the A i are disjoint measurable sets. ( Comment : It might seem that finite additivity implies the condition (M1): we have ( ) = ( ) = ( ) + ( ), and so ( ) = 0. What is wrong with this argument? Under what additional condition on would finite additivity imply (M1) ?) Exercise . If A and B are measurable sets with A B , show that ( A ) ( B ). (Hint : write B as the union of A and B \ A ; check that B \ A (which is B A c ) is measurable and use additivity.) The simplest meaningful example of a measure is counting measure...
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## This note was uploaded on 01/27/2010 for the course MATH 7312 taught by Professor Sengupta during the Spring '02 term at LSU.

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msreinteg - Lecture 2. Measure and Integration There are...

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