hmwk12sol_a

hmwk12sol_a - MATH 202A HOMEWORK 12 Exercise 4 Proof Let...

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MATH 202A HOMEWORK 12 Exercise 4 . Proof. Let ˜ g ( x ) = R b 0 χ { x t } | f ( t ) | t dt . It is clear that the function in the integral is mea- surable and thus by Tonelli Theorem, we have R b 0 ˜ g ( x ) = R b 0 ± R b 0 χ { x t } | f ( t ) | t dt ² dx = R b 0 ± R b 0 χ { x t } | f ( t ) | t dx ² dt = R b 0 | f ( t ) | t ± R b 0 χ { x t } dx ² dt = R b 0 | f ( t ) | t ± R t 0 dx ² dt = R b 0 | f ( t ) | t t dt = R b 0 | f ( t ) | dt < . Hence, ˜ g is integrable and thus g ( x ) is also integrable. Now we can use Fubini Theorem and similarly, by replacing | f ( t ) | with f ( t ), we have R b 0 g ( x ) dx = R b 0 f ( t ) dt . ± Exercise 5 . (a) Proof. Since F is closed in R , x 0 ,y 0 , s.t. | x - x 0 | = δ ( x ) , | y - y 0 | = δ ( y ). Hence, | x - x 0 | ≤ | x - y 0 | , | y - y 0 | ≤ | y - x 0 | . By triangle inequality, we have | x - y 0 | = | x - y + y - y 0 | ≤ | x - y | + | y - y 0 | and thus | x - x 0 |-| y - y 0 | ≤ | x - y 0 |-| y - y 0 | ≤ | x - y | . Similarly, we have | y - y 0 |-| x - x 0 | ≤ | x - y | . Therefore, we know | δ ( x ) - δ ( y ) | = || x - x 0 | - | y - y 0 || ≤ | x - y | . ± (b) Proof. F c = t n N U n , where U n are disjoint open sets and the expression is unique. x F c , suppose x ( a,b ). Let I = ( x + a 2 , x + b 2 ). Hence, y I,δ ( y ) min { a - x 2 , b - x 2 } , r . Thus, we have I ( x ) R I δ ( y ) | x - y | 2 dy R I r | x - y | 2 dy r R ( a - x 2 , b - x 2 ) 1 z 2 dz . Since 0 ( a - x 2 , b - x 2 ), we have R ( a - x 2 , b - x 2 ) 1 z 2 dz = and thus I ( x ) ≥ ∞ , which implies I ( x ) = , x F c . ± (c) Proof. By hint, we investigate R F I ( x ) dx . Since δ ( x ) is continuous, δ ( y ) | x - y | 2 is also continuous and thus measurable. Hence, I ( x ) is non-negative and measurable function. By extended Fubini Theorem and part (a), we have R F I ( x ) dx = R F R F c δ ( y ) | x - y | 2 dydx = R
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hmwk12sol_a - MATH 202A HOMEWORK 12 Exercise 4 Proof Let...

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