# Solution yes since f is continuous on a b so is f

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Solution: Yes, since f is continuous on [ a, b ], so is | f | , hence, by theorem 5.10, | f | is integrable on [ a, b ]. 6. (20 pts) Suppose f : [ a, b ] R is continuous and increasing. Prove that sup f ( E ) = f (sup E ) for every nonempty set E [ a, b ]. Solution: Let E [ a, b ]. Thus E is bounded, so s = sup E is finite and s [ a, b ]. Because f is increasing, for any x [ a, b ], we have f ( a ) f ( x ) f ( b ). In particular f ( E ) is bounded, so sup f ( E ) is finite. Again because f is increasing and s x, x E , we have f ( s ) f ( x ) x E . Thus, f ( s ) is an upper bound for f ( E ), so we get f (sup E ) sup f ( E ). To obtain the other inequality, let x n be a sequence of elements from E such that x n sup E (such a sequence exists, by the approximation property of the supremum). Then sup f ( E ) f ( x n ), for any n N (because x n E ). Taking the limit on n in this inequality and using the fact that f is continuous, we get sup f ( E ) f ( s ) = f (sup E ).
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