PMATH450_S2015.pdf

# R f g and r φ n r n r f r g the result holds note we

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R f + g and R φ n + R n ! R f + R g , the result holds. Note. We must still check that R f + g = R f + R g for all integrable functions. This is left as an exercise. Example. For measurable E , Z R f = Z f ( χ E + χ E C ) = Z f χ E + Z f χ E C = Z E f + Z E C f. Lemma (Fatou’s Lemma) . Let f n 0 be measruable. Then R lim inf f n lim inf R f n . Proof. Let g n = inf k n f k , so g n f k 8 k n . By monotonicity, R g n R f k 8 k n , so R g n inf k n R f k . Also g n % . Hence lim n !1 Z g n lim n !1 inf k n Z f k = lim inf Z f n . But g n % lim inf f n . By MCT, R g n ! R lim inf f n . Thus R lim inf f n = lim R g n lim inf R f n . Lecture 10: May 27 Corollary. 1. If f n 0 are measurable and f n ! F , then Z f = Z lim inf f n lim inf Z f n . 2. If f n are measurable and f n ! f with R | f n | 1 8 n . Then | f n | ! | f | and Z | f | lim inf Z | f n | 1 , so f is integrable. Theorem (Dominated Convergence Theorem) . Suppose f n are measurable and f n ! f pointwise and there exists an integrable function g such that | f n ( x ) | g ( x ) 8 n, x . Then R f n ! R f . In fact, R | f n - f | ! 0. Proof. Look at 2 g - | f n - f | , which is measurable. Also g | f n | ! | f | , so | f n - f | | f n | + | f | 2 g . Thus 2 g - | f n - f | 0, so we can apply Fatou’s lemma: Z lim inf(2 g - | f n - f | ) lim inf Z (2 g - | f n - f | ) Notice that lim inf(2 g - | f n - f | ) = 2 g since f n ! f . Thus Z 2 g lim inf Z (2 g - | f n - f | ) = lim inf ✓Z 2 g - Z | f n - f | = Z 2 g - lim sup Z | f n - f | Since g is integrable, R 2 g < 1 , so we can subtract it o both sides to get lim sup R | f n - f | 0. Then 0 lim inf Z | f n - f | lim sup Z | f n - f | 0

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2 INTRODUCTION TO LEBESGUE INTEGRATION 16 Thus lim R | f n - f | = 0. To see that R f n ! R f , first note that f n and f are indeed integrable, so we can consider Z ( f n - f ) Z | f n - f | ! 0 so R ( f n - f ) ! 0 and thus R f n ! R f . Example. 1. 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 . Any g f n for all n will have to be n on (0 , 1 n ), i.e. g n on h 1 n +1 , 1 n for all n . Then g G = P 1 n =1 n χ [ 1 n +1 , 1 n ) . Let S k be the k th partial sum. Each S k is simple and positive, and S k % G pointwise. By MCT, Z G = lim Z S k = lim Z k X n =1 n χ [ 1 n +1 , 1 n ) = lim k X n =1 Z n χ [ 1 n +1 , 1 n ) = lim k X n =1 n 1 n - 1 n + 1 = lim k X n =1 1 n + 1 = 1 So G is not integrable, and hence neither is g . 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 . This fails the DCT because the smallest function that dominates f n will be 1 on [1 , 1 ], and thus not integrable. Note. If f n 0 and measurable and g = P 1 n =1 f n , then by MCT, since P k n =1 f n % g , Z g = lim k !1 Z k X n =1 f n = lim k !1 k X n =1 Z f n = 1 X n =1 Z f n . Thus the integral is countably additive for nonnegative functions. Cantor Set. Defined recursively by starting with C 0 = [0 , 1] and removing the middle third of each closed interval in C n to get C n +1 .
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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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