PMATH450_S2015.pdf

7 for fixed y r r f x y dm x r r f x dm x due to the

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7. For fixed y , R R f ( x + y ) dm ( x ) = R R f ( x ) dm ( x ), due to the translation invariance of m . Proof. Suppose f = P a k χ E k . Then Z X a k χ E k ( x + y ) dm ( x ) = Z X a k χ E k - y ( x ) dm ( x ) = X a k m ( E K - y ) = X a k m ( E k ) = Z X a k χ E k = Z f. Let g y ( x ) = g ( x + y ). If f 0, then φ f i φ y f y . If f is real then ( f ± ) y = f ± y . If f is complex, then Re( f ) y = Re( f y ) and Im( f ) y = Im( f y ). Lecture 9: May 25

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2 INTRODUCTION TO LEBESGUE INTEGRATION 14 2.5 Convergence Theorems Example. 1. If f n is measurable and f n ! f , it is not necessarily true that R f n ! R f . Let f n ( x ) be n for x 2 (0 , 1 n ) and 0 otherwise. That is f n = n χ (0 , 1 n ) . Note that f n ! 0 pointwise, but R [0 , 1] f n = 1 for all n . 2. Let f n = 1 on [ n, n + 1] and 0 otherwise. Then f n ! 0, but R R f n = 1 for all n . Theorem (Monotone Convergence Theorem) . Suppose f n 0 (allowing 1 ) are measurable functions with f n ( x ) % f ( x ) for all x . Then R E f = lim n !1 R E f ( n ) for all measurable E . Note. Note that this fails for Riemann integrability: Let Q \ [0 , 1] = { r n } 1 1 . Define f n ( x ) = 1 if x = r 1 , . . . , r n and 0 otherwise. f n 0 an measurable for all n , and f n ( x ) % f = χ Q \ [0 , 1] . Then R R 1 0 f n = 0, but R R 1 0 f does not exist. However, R [0 , 1] f n = 0 = R [0 , 1] f . Lemma. Let φ be positive and simple, A n be measurable sets with A n A n +1 for all n . Put A = S 1 n =1 A n . Then R A n φ ! R A φ , or equivalently R φχ A n ! R φχ A . Proof. This is a special case of MCT with f n = φχ A n and f = φχ A . Let φ = P N i =1 a i χ E i . Then Z A n φ = N X i =1 a i m ( A n \ E i ) . Note that E i \ A n E i \ A n +1 8 n and S 1 n =1 E i \ A n = E i \ A . By continuity of measure, m ( E i \ A n ) ! m ( E i \ A ) for each i . Hence N X i =1 a i m ( E i \ A n ) ! N X i =1 a i m ( E i \ A ) = Z A φ . Proof of MCT. Recall that f n measurable, f n ! f implies f measurable. Since f n f , by monotonicity, R E f n R E f . lim n !1 R E f n exists by monotonicity and lim n !1 R E f n R E f . We must prove that R E f lim n !1 R E f n . Recall that R E f = sup 0 φ f φ simple R E φ , thus it will be enough to prove that R φ lim n !1 R f n for all simple φ with 0 φ f . Fix such a φ , and take any 0 < < 1. Define A n = { x 2 E : f n ( x ) ↵φ ( x ) } = E \ ( f n - ↵φ ) - 1 [0 , 1 ], which are measurable. Also note that A n A n +1 since the f n s are increasing. Now we claim that S A n = E . Let x 2 E . If φ ( x ) = 0 then f n ( x ) ↵φ ( x ) 8 n , so x 2 A n 8 n . Otherwise φ ( x ) > 0, in which case ↵φ ( x ) < φ ( x ) f ( x ). Since f n ( x ) ! f ( x ), for large enough n we have f n ( x ) > ↵φ ( x ), so x 2 A n , and hence x 2 S A n . Hence E S A n , and clearly S A n E , so S A n = E . Now by the lemma, Z A n φ ! Z S A j φ = Z E φ . On the other hand, Z A n φ = Z A n ↵φ Z A n f n = Z f n χ A n Z f n χ E = Z E f n ! lim n !1 Z E f n So R E φ lim n !1 R E f n for every 2 (0 , 1). Let ! 1 to get R E φ lim R E f n . Proposition. If f, g 0 are measurable, then R f + g = R f + R g .
2 INTRODUCTION TO LEBESGUE INTEGRATION 15 Proof. We have already proved this for simple functions. Get simple functions 0 φ n % f and 0 n % g . Then φ n + n % f + g . We have R φ n + n = R φ n + R n . Since R φ n + n !

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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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