Notice that sup f v z inf f v z 1 k look at v z z 2 e

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Notice that sup f V z - inf f V z 1 k < " . Look at { V z : z 2 E C k } and note that S z 2 E C k V z E C k X . Since X is compact and each V z is open, find a finite subcover V 1 , . . . , V m of X . Now I 1 , . . . , I n , V 1 , . . . , V m covers [ a, b ]. Take as a partition P the endpoints of these intervals, along with the points a and b . For each pair of consecutive points x i - 1 , x i in the partition, we either have x i - 1 , x i 2 I j or x i - 1 , x i 2 V j . If x i - 1 , x i 2 V j , then M i - m i sup f V j - inf f V j < " and if x i - 1 , x i 2 I j , then M i - m i 2 C , where | f ( x ) | C for all x 2 [ a, b ]. Now we have U ( f, P ) - L ( f, P ) = X ( M i - m i )( x i - x i - 1 ) = X x i - 1 ,x i 2 V j for some j ( M i - m i )( x i - x i - 1 ) + X x i - 1 ,x i 2 I j for some j x i - 1 ,x i / 2 V j for any V j ( M i - m i )( x i - x i - 1 ) < " X ( x i - x i - 1 ) + 2 C X x i - 1 ,x i 2 I j for some j x i - 1 ,x i / 2 V j for any V j ( x i - x i - 1 ) " ( b - a ) + 2 C n X j =1 ` ( I j ) = " ( b - a ) + 2 C " = " ( b - a + 2 C ) Since b - a + 2 C is a constant, we could get this arbitrarily small by choosing a su ffi ciently small " . Conversely, suppose f is Riemann integrable, and we want to prove m ( E ) = 0. Since E = S 1 k =1 E k , it is enough to show that m ( E k ) = 0 for each k . So fix k and " > 0. Since f is Riemann integrable, we can find a partition P such that U ( f, P ) - L ( f, P ) < " k . Let I = { i : ( x i - 1 , x i ) \ E k 6 = ; } . Now " k > U ( f, P ) - L ( f, P ) = X ( M i - m i )( x i - x i - 1 ) X i 2 I ( M i - m i )( x i - x i - 1 ) X i 2 I 1 k ( x i - x i - 1 )
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2 INTRODUCTION TO LEBESGUE INTEGRATION 10 so P i 2 I ( x i - x i - 1 ) < " . Then E j S i 2 I ( x i - 1 , x i ) [ { x j } n j =0 , so m ( E k ) X i 2 I m ( x i - 1 , x i ) + m { x j } n j =0 = X i 2 I ( x i - x i - 1 ) + 0 < " . Lecture 7: May 20 2.3 Measurable Functions Definition. A real-valued function f is called Lebesgue meausrable if 8 2 R , { x : f ( x ) < } is Lebesgue measurable. A complex-valued function f is measurable if Re( f ) and Im( f ) are measurable. Example. (1) f ( x ) = c is measurable. (2) Continuous functions are measurable. Since f - 1 ( -1 , ) is open for continuous functions, and hence Lebesgue measurable. Note. In the definition of measurability of a function, we could replace { x : f ( x ) < } with { x : f ( x ) > } , { x : f ( x ) } , or { x : f ( x ) } . Proof. Observe that { x : f ( x ) < } = { x : f ( x ) } C , so f is measurable i all { x : f ( x ) } are measurable. If f is measurable, then S 1 n =1 { x : f ( x ) + 1 n } are measurable for all so { x : f ( x ) > } is equivalent. Similarly, { x : f ( x ) } = { x : f ( x ) > } C is equivalent. Definition. The characteristic function for a set E is the function χ E defined by χ E ( x ) = ( 1 x 2 E 0 x / 2 E . Note. { x : χ E ( x ) < } = 8 > < > : ; 0 E C 0 < 1 R > 1 so χ E is measurable i E is measurable. Definition. A simple function is a function of the form f = P N i =1 a i χ E i where each E i is measurable and each a i 2 R . Note. The simple functions are the measurable functions that take on finitely many values. Example. Step functions are simple. Proposition. If f and g are measurable, then so are f ± g , fg , and f g where g 6 = 0.
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