Now let q 0 1 2 r i 1 1 then e r i 0 1 since e 0 1 2

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Now let Q \ [0 , 1 2 ] = { r i } 1 1 . Then E + r i [0 , 1] since E [0 , 1 2 ]. Then m 1 [ i =1 E + r i ! = 1 X i =1 m ( E + r i ) = 1 X i =1 m ( E ) = 1 but S 1 i =1 E + r i [0 , 1], so by monotonicity and property 1, m ( S 1 i =1 E + r i ) m ([0 , 1]) = 1. So we cannot have a measure function with all of the desired properties.
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2 INTRODUCTION TO LEBESGUE INTEGRATION 6 2.1 Construction of Lebesgue Measure Definition. Let A R and let C ( A ) = {{ I n } 1 n =1 : I n open interval with S 1 n =1 I n A } . We define the outer Lebesgue measure as m ( A ) = inf ( 1 X n =1 ` ( I n ) : { I n } 2 C ( A ) ) . Properties. (1) m ( ; ) = 0. (2) m ( { x } ) = 0 for all x 2 R . (3) If A B then m ( A ) m ( B ). (4) m is translation invariant. Proposition. The outer measure of an interval is its length. Lecture 4: May 11 Proof. Case: [ a, b ]. If " > 0 then [ a, b ] ( a - " , b + " ). Thus m ([ a, b ]) ` (( a - " , b + " )) = b - a +2 " . Since this is true for all " > 0, so m ([ a, b ]) b - a . Now we claim that whenever { I n } 2 C [ a, b ], we have P ` ( I n ) b - a . Let { I n } 1 n =1 2 C [ a, b ], so by compactness of [ a, b ] there exists a finite subcover { I j 1 , . . . , I j m } . Then P 1 i =1 ` ( I i ) P m k =1 ` ( I j k ). So it is enough to show that whenever I 1 , . . . , I n is a cover of [ a, b ] by open intervals, then P n j =1 ` ( I j ) b - a . Since S n i =1 I j [ a, b ], pick j 1 such that a 2 I j 1 = ( a 1 , b 1 ), so a 1 < a < b 1 . If b < b 1 then [ a, b ] ( a 1 , b 1 ), so P n j =1 ` ( I j ) ` ( I j 1 ) b 1 - a 1 > b - a and we are done. Otherwise a < b 1 b , so b 1 2 I j 2 = ( a 2 , b 2 ) for some j 1 6 = j 2 . If b < b 2 , then [ a, b ] ( a 1 , b 1 ) [ ( a 2 , b 2 ) and P n i =1 ` ( I j ) b 2 - a 2 + b 1 - a 1 b 2 - a 1 b - a . Otherwise we repeat the argument. Get intervals I j k = ( a k , b k ) with a k < b k - 1 < b k where the j k are all distinct. If b < b k we have [ a, b ] S k i =1 I j i and then P k i =1 ` ( I j i ) = P k i =1 b i - a i b - a . This process must terminate after finitely many steps because there are only finitely many I j s to pick from. The remaining cases are left as an exercise. Proposition. m 1 [ k =1 A k ! 1 X k =1 m ( A k ) 8 A k R Proof. Fix " > 0. For each A k , choose open intervals { I n,k } 1 n =1 that cover A k such that P 1 n =1 ` ( I n,k ) m ( A n ) + " 2 k . Then { I n,k } 1 n =1 1 k =1 2 C ( S 1 n =1 A k ), so m ( S 1 k =1 A k ) P 1 k =1 P 1 n =1 ` ( I n,k ) P 1 k =1 ( m ( A k ) + " 2 k ) = P 1 k =1 m ( A k ) + " . But " > 0 was arbitrary, therefore m ( S 1 k =1 A k ) P 1 k =1 m ( A k ). Corollary. Let A = { x k } 1 k =1 be countable. Then m ( A ) = 0. Proof. m ( A ) = m ( { x k } 1 k =1 ) P 1 k =1 m ( { x k } ) = 0. Corollary. For all A R and " > 0, there exists an open set O A such that m (0) m ( A ) + " . Proof. Choose intervals { I n } covering A such that P 1 n =1 ` ( I n ) m ( A ) + " . Take O = S 1 n =1 I n A , which is open. We have m ( O ) = m ( S 1 n =1 I n ) P 1 n =1 m ( I n ) = P n =1 ` ( I n ) m ( A ) + " . Definition. We say that A R is Lebesgue measurable if whenever E R , m ( E ) = m ( E \ A ) + m ( E \ A C ).
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2 INTRODUCTION TO LEBESGUE INTEGRATION 7 Example. A = ; and A = R are both Lebesgue measurable.
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