Proof let f x φ x 6 0 so m f 0 then φ 0 χ f c p n

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Proof. Let F = { x : φ ( x ) 6 = 0 } so m ( F ) = 0. Then φ = 0 χ F C + P N k =1 A k F k where F k F . Then Z E φ = 0 m ( E \ F C ) + N X k =1 a k m ( E \ F k ) | {z } 0 = 0 + 0 = 0 . Definition (Lebesgue Integral for Non-negative Real-Valued Functions) . Let f : R ! [0 , 1 ] be measurable. Let E be measurable. Then we define Z E f dm = Z E f = sup 0 φ f φ simple Z E φ Note that this is consistent with our previous definition for the Lebesgue integral for simple functions. Note. If f g then R E f R E g . Definition (Lebesgue Integral for Real-Valued Functions) . Let f : R ! R be measurable. Let E be measurable. Let f + ( x ) = ( f ( x ) f ( x ) 0 0 otherwise = max( f, 0) and f - ( x ) = ( - f ( x ) f ( x ) < 0 0 otherwise = min( f, 0) These are both measurable, and f = f + - f - . We now define the Lebesgue integral of f on E as R E f = R E f + - R E f - provided this is not 1 - 1 . We say that f is integrable if R E | f | < 1 . Note. Note that | f | = f + + f - , so if f is measurable, then so are f + and f - , and thus so is | f | . Note. Since f + , f - | f | , monotonicity of the integral implies that R E f + < 1 and R E f - < 1 , so R E f is well- defined. Definition (Lebesgue Integral for Complex-Valued Functions) . Suppose f : R ! C is measurable and E is measur- able. We say f is integrable if both Re( f ) and Im( f ) are integrable, and define R E f = R E Re( f ) + i R E Im( f ). Properties of the Lebesgue Integral. 1. If f is measurable and bounded by M , then R E | f | R E M = Mm ( E ). If E = [ a, b ], then R E | f | M ( b - a ) < 1 , so f is integrable over E . 2. R E f = R R f χ E , which is well-defined, since f χ E is measurable when f, E are. Proof. First suppose f = P a k χ E k is simple. Then Z E f = X a k M ( E K \ E ) = Z R X a k χ E k \ E = Z R X a k χ E k χ E = Z R f χ E . Now suppose f : R ! [0 , 1 ]. Then Z E f = sup 0 φ f φ simple Z E φ = sup 0 φ f φ simple Z R φχ E
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2 INTRODUCTION TO LEBESGUE INTEGRATION 13 Now φ f ) φχ E f χ E and if f χ E is simple, then = 0 χ E where 0 = on E and 0 otherwise. Note that 0 is simple, so = 0 χ E f χ E , so sup 0 φ f φ simple Z R φχ E = sup 0 f χ E simple Z R = Z R f χ E . For f : R ! R , note that f ± χ E = ( f χ E ) ± . For f : R ! C , Re( f ) χ E = Re( f χ E ) and Im( f ) χ E = Im( f χ E ) and these observations lead to the completion of the proof. 3. R E f = 0 if m ( E ) = 0. Proof. If f is simple, it’s clear. If f 0 then R E f = sup 0 φ f φ simple R E φ = 0. If f : R ! R , then R E f = R E f + + R E f - = 0 - 0 = 0. The complex case is similar. 4. R E f = R E f for any scalar . 5. | R E f | R E | f | . Proof. For some | | = 1, we have Z f = Z f = Z f = Z Re( f ) + i Z Im( f ) | {z } 0 = Z Re( f ) Z | f | = Z | f | . 6. If φ and are simple, then R φ + = R φ + R . Proof. Say φ = P a k χ E k and = P b j χ F j where S E k = R = S F j . Let G i be the sets of the form E k \ F j . Then S j E k \ F j = E k , so S G i = S E k = R . If x 2 G i = E k \ F j , then φ + = a k + b j . Denote φ = P i χ G i and = P β i χ G i , so φ + = P ( i + β i ) χ G i . Then Z φ + = X ( i + β i ) m ( G i ) = Z φ + Z .
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