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hw5 - Math 245A Homework 5 Brett Hemenway 21 Suppose fn f...

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Math 245A Homework 5 Brett Hemenway November 16, 2005 21. Suppose f n , f L 1 , and f n f a.e. Recall || f n | - | f || ≤ | f n - f | which means 0 | f n | - | f | = ( | f n | - | f | ) || f n | - | f || | f n - f | So if | f n - f | → 0, then | f n | - | f | → 0, so | f n | → | f | . We also have | f n - f | ≤ | f n | + | f | So if | f n | → | f | , we have ( | f n | + | f | ) 2 | f | . Then, using problem 20, we have | f n - f | → | f - f | = 0 Because | f n - f | → 0 in L 1 . This gives | f n - f | → 0 iff | f n | → | f | . 23. Let f : [ a, b ] R be a bounded function and H ( x ) = lim δ 0 sup | y - x |≤ δ f ( y ) h ( x ) = lim δ 0 inf | y - x |≤ δ f ( y ) 1
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a. If f is continuous at x , then for any > 0, there exists a δ , such that | x - y | ≤ δ implies | f ( x ) - f ( y ) | < , so f ( x ) - inf | y - x |≤ δ f ( y ) sup | y - x |≤ δ f ( y ) f ( x ) + Which means H ( x ) = lim δ 0 sup | y - x |≤ δ f ( y ) = lim δ 0 inf | y - x |≤ δ f ( y ) = h ( x ) Conversely, if H ( x ) = h ( x ), then for any > 0, there exists a δ such that sup | y - x |≤ δ f ( y ) - inf | y - x |≤ δ f ( y ) < so we have | f ( x ) - f ( y ) | < whenever | y - x | ≤ δ b. If { P k } is sequence of ever finer partitions, and G p , g p are defined as before, G P = n j =1 M j χ ( t j - 1 ,t j ] g P = n j =1 m j χ ( t j - 1 ,t j ] with G = lim G P , g = lim g P . We have, G P k ( x ) H ( x ) for all x outside of P k g P k ( x ) h ( x ) for all x outside of P k because for any x ( t j - 1 , t j ) we can find a δ such that ( x - δ, x + δ ) ( t j - 1 , t j ). Thus for x k =1 P k we have G P k ( x ) H ( x ) g P k ( x ) h ( x ) So outside of k =1 P k G ( x ) H ( x ) g ( x ) h ( x ) 2
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Now, let H δ ( x ) = sup | y - x |≤ δ f ( y ) h δ ( x ) = inf | y - x |≤ δ f ( y ) Then choose a k 0 such that every interval ( t j - 1 , t j ) in P k has t j - t j - 1 < δ , then G P k ( x ) H δ ( x ) g P k ( x ) h δ ( x ) holds for all k > k 0 , so G ( x ) H δ ( x ) g ( x ) h δ ( x ) for any δ , so G ( x ) H ( x ) g ( x ) h ( x ) So we have that G ( x ) = H ( x ) outside of k =1 P k g ( x ) = h ( x ) outside of k =1 P k but each P k is finite, so k =1 P k is countable, and hence a null set. Then since G, g are Lebesgue measurable we have that H, h are also Lebesgue measurable, and [ a,b ] Hdm = [ a,b ] Gdm = I b a ( f ) [ a,b ] hdm = [ a,b ] gdm = I b a ( f ) By part (a), [ a,b ] Hdm = [ a,b ] hdm iff the set of discontinuities of f in [ a, b ] has Lebesgue measure zero. So by the previous
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