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Unformatted text preview: 2 Lebesgue integration 1. Let ( , A , ) be a measure space. We will always assume that is com plete, otherwise we first take its completion. The example to have in mind is the Lebesgue measure on R n , ( R n , L n ,   ) . We will build the inte gration theory for Ameasurable functions. We will consider measurable functions f :  R , where R = R 1 {} { + } (and also functions F : C {} , where C = C {} ). First we define integrals of real valued nonnegative functions, and then reduce the general case to this. We will follow Rudin very closely. 2. Let f : R be a simple function : f = N X j =1 c j E j , c 1 ,...,N R , E 1 ,...,N A , N < + . It is useful to know that any simple f can be written as a linear com bination of E j , E j A , with distinct c 1 , ... , c N and disjoint E 1 , ... , E N . In fact, a simple f takes only a finite number of distinct values 1 < 2 < < M . Hence the desired representation is f = M X m =1 m f 1 ( { m } ) . Adding a term with c j = 0 if necessary, we can always assume that N [ j =1 E j = . 3. Let f be a simple function written as f = N X n =1 c n E n , E j E k = if j 6 = k. Define the Lebesgue integral of f with respect to over by Z f ( x ) d ( x ) = Z f d def = N X j =1 c j ( E j ) agreeing that = 0 . 1 We do not assume here that c 1 ,...,N are distinct. (However, since E j s are disjoint, we have c 1 ,...,N .) The definition is correct. In fact, suppose we also have f = M X m =1 d m F m , F j F k = if j 6 = k. Notice that A B = A + B if A B = . But then f = N X n =1 M X m =1 e nm E n F m , e nm = c n = d m . Now, keeping in mind that all sets E n (and all F m ) are disjoint we derive N X n =1 c n ( E n ) = N X n =1 c n M X m =1 ( E n F m ) = N X n =1 M X m =1 e nm ( E n F m ) , and by the same argument M X m =1 d m ( F m ) = N X n =1 M X m =1 e nm ( E n F m ) . 4. For a general measurable F define Z F ( x ) d ( x ) = Z F d = sup f simple , f F Z f d. The definition is correct when applied to simple functions. This means, that for any f , f = N X n =1 c n E n with disjoint E 1 , ... , E N , the equality sup simple , f Z d = N X n =1 c n ( E n ) holds. In fact, take any simple written as = M X m =1 m F m 2 with disjoint F 1 , ... , F M , such that f . Then for all m,n m F m E n c n F m E n . Keeping in mind that all sets E n (and all F m ) are disjoint we derive M X m =1 m ( F m ) = M X m =1 N X n =1 m ( F m E n ) M X m =1 N X n =1 c n ( F m E n ) = N X n =1 M X m =1 c n ( F m E n ) N X n =1 c n ( E n ) ....
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 Fall '07
 RickRugangYe

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