6 σ subadditivity if a n 2 m then s 1 n 1 a n 2 m

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(6) σ -subadditivity: If A n 2 M , then S 1 n =1 A n 2 M and m ( S 1 n =1 A n ) P 1 n =1 m ( A n ). In fact, the Lebesgue measure of a countable set is 0. (7) Continuity of Measure: If A 1 A 2 · · · S 1 n =1 A n = A with each A n 2 M , then A 2 M and m ( A ) = lim n !1 m ( A n ) = sup n 2 N m ( A n ). Proof. Let B 1 = A 1 , and B n = A n \ A n - 1 for n > 1. Then B n 2 M for each n and the B n s are pairwise disjoint, with S 1 n =1 B n = A . By σ -additivity, m ( A ) = 1 X n =1 m ( B n ) = lim N !1 N X n =1 m ( B n ) = lim N !1 m N [ n =1 B n ! = lim N !1 m ( A N ) . (8) Downward Continuity of Measure: If A 1 A 2 · · · T 1 n =1 A n = A with each A n 2 M , then A 2 M and m ( A ) = lim n !1 m ( A n ) provided m ( A 1 ) < 1 . Proof. Let B n = A 1 \ A n and let B = A 1 \ A . Then B n B n +1 for each n , and S 1 n =1 B n = B . By continuity of measure, m ( A 1 ) - m ( A ) = m ( A 1 \ A ) = m ( B ) = lim n !1 m ( B n ) = lim n !1 m ( A 1 \ A n ) = lim n !1 [ m ( A 1 ) - m ( A n )] = m ( A 1 ) - lim n !1 m ( A n ) And thus m ( A ) = lim n !1 m ( A n ). Lecture 6: May 15 2.2 Characterization of Riemann Integrability Theorem. A bounded function f : [ a, b ] ! R is Riemann integrable over [ a, b ] if and only if m { discontinuities of f } = 0. Recall. m ( E ) = 0 i m ( E ) = 0 i 8 " > 0 9 a collection of open intervals { I n } 1 n =1 that cover E with P ` ( I n ) < " . Proof of Theorem 11. Let E denote the set of points at which f is discontinuous. Let E k = z 2 [ a, b ] : 8 δ > 0 9 x, y 2 [ z - δ , z + δ ] such that | f ( x ) - f ( y ) | 1 k .
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2 INTRODUCTION TO LEBESGUE INTEGRATION 9 If z 2 E k then z 2 E . Indeed, if f was continuous at z , then for " = 1 2 k , 9 δ > 0 such that 8 x 2 [ z - δ , z + δ ], | f ( x ) - f ( z ) | < " = 1 2 k . Then if x, y 2 [ z - δ , z + δ ], then | f ( x ) - f ( y ) | | f ( x ) - f ( z ) | + | f ( z ) - f ( y ) | < 2 1 2 k = 1 k , so z / 2 E k . Therefore S 1 k =1 E k E . In fact, if z / 2 E k , there exists δ > 0 such that 8 x, y 2 [ z - δ , z + δ ], we have | f ( x ) - f ( y ) | < 1 k . If we let y = z , then | f ( x ) - f ( z ) | < 1 k . So if z / 2 S E k , for each " > 0 we can pick k such that 1 k < " . z / 2 E k , so if δ is chosen as above and x 2 [ z - δ , z + δ ], then | f ( x ) - f ( z ) | < 1 k < " , so f is continuous at z , and thus z / 2 E . Hence we can conclude that E = S 1 k =1 E k . Side Note (Not part of proof) : We claim now that E k is closed, which will prove that E is F σ , and hence Borel. Let ( z n ) be a sequence in E k with z n ! z . Fix δ > 0, and get z n 2 [ z - δ 2 , z + δ 2 ]. Since z n 2 E k , there exist points x, y 2 [ z n - δ 2 , z n + δ 2 ] [ z - δ , z + δ ] such that | f ( x ) - f ( y ) | 1 k . Thus z 2 E k so E k is closed. Suppose m ( E ) = 0. Let " > 0 and find a partition P of [ a, b ] such that U ( f, P ) - L ( f, P ) < " . Denote Equivalently, X sup f [ x i - 1 ,x i ] - inf f [ x i - 1 ,x i ] ( x i - x i - 1 ) < " Denote M i = sup f [ x i - 1 ,x i ] and m i = inf f [ x i - 1 ,x i ] . Since m ( E ) = 0, we can pick open intervals { I n } 1 n =1 such that S 1 n =1 I n E and P 1 n =1 ` ( I n ) < " . Pick k such that 1 k < " . Note that S 1 n =1 I n E E k and E k is closed and E k [ a, b ], so E k is compact. By compactness, take a finite subcover I 1 , . . . , I n of E k . Let X = [ a, b ] \ S n j =1 I j . If z 2 Z , then z / 2 S n j =1 I j E k , so z / 2 E k . Thus there exists δ > 0 such that 8 x, y 2 [ z - δ , z + δ ], we have | f ( x ) - f ( y ) | < 1 k . Let V z = ( z - δ , z + δ ) and V z = [ z - δ , z + δ ].
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