1873_solutions

# Least upper bound of this set we have m un me n1 3

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Unformatted text preview: e on a, b we shall show that f satisfies the second criterion for integrability. Suppose that  0. Using the given property of f we choose an elementary subset A of a, b such that m A  /2 and such that the function f 1  A is integrable on a, b . Choose a partition P 1 of a, b such that the function  A steps within P. Now, using the fact that f 1  A satisfies the second criterion for integrability we choose a partition P 2 of a, b such that if we define B x a, b w P, f 1  A x . then m B  /2. We now define P to be the common refinement of P 1 and P 2 and we express P as P  x 0 , x 1 , , x n . For each j  1, 2, , n, if the open interval x j 1 , x j is not included in A Þ B then, since the functions f and f 1  A agree in the interval x j 1 , x j the condition must hold whenever x w P, f x  w P, f 1  A x j 1 , x j . Since m AÞB m A m B  x   2 2 we have succeeded in showing that f satisfies the second criterion for integrability. 5. a. Suppose that f is a nonnegative function defined on an interval a, b and that for every number set x a, b fx  0, the is finite. Prove that f must be integrable on a, b and that Þ f  0. a We shall show that f satisfies the first criterion for integrability. Suppose that  0. Choose a finite set S such that fx  ba whenever x a, b S and define P to be the partition of a, b whose points are the numbers a and b and the members of S arranged in increasing order. Since l P, f is nonnegative and u P, f never exceeds the value / b a , we have b Þ a w P, f b  Þ a u P, f b Þ a u P, f b l P, f . Since f satisfies the first criterion for integrability, f is integrable on a, b and the same b b from which we deduce that Þ f  0. argument shows that whenever  0 we have Þ f a a b. Prove that if f is the ruler function that was introduced in an earlier example then f is an integrable function on the interval 0, 1 , even though f is discontinuous at every rational number in the interval. The ruler function obviously has the property described in part a. 6. Given that f is a bounded nonnegative function defined on an interval a, b , prove that the following conditions are equivalent: a. The function f is integrable on the interval a, b and Þ f  0. b a b. For every number  0 there exists an elementary set E such that m E  x a, b fx E. and such that To show that condition a implies condition b we assume that f is integrable on a, b . Suppose that  0. To obtain the desired set E we shall use the same sort of technique as was used in the proof of Theorem 11.8.4. Choose a step function S f such that 273 Þa S  b 2 and define E x a, b . Sx We observe that x a, b fx E. Now since S is a step function, the set E is elementary and we have  2 Þ a S ÞE S ÞE b  mE from which we deduce that m E  . To show that condition b implies condition a we assume that condition b holds. Once again we borrow from the proof of proof of Theorem 11.8.4. Using the fact that f is bounded we choose a number k such that f x  k fo...
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