3.18 Suppose if x = 0, f(x) = if z e (. where n e N. Prove: f € R[0, 1], even though f has infinitely many discontinuities. (Hint: Consider Theorem 3.2.3.)

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3.18 Suppose
o if r = 0,
f(x) =
t if ze ( .
where n e N. Prove: f € R[0, 1], even though f has infinitely many discontinuities.
(Hint: Consider Theorem 3.2.3.)
For reference
Theorem 3.2.3 If f is monotone on [a, b), then f € R[a, b).
Proof: Suppose first that f is increasing. Then, for each partition P of (a, b] we
have
U(f, P) – L(f, P) =Ei (#+) – f(Ti-1)]A¤4
k=1
i=1
= ||P||S() – f(a)] → 0
as ||P|| → 0. Thus f € R(a, b). Finally if f is decreasing, then - f is integrable by
the first part of this proof, and so f = -(-f) € R{a, b] as well.
Here is a variant of the Darboux Integrability Criterion which is often useful. (See
Exercises 3.26 and 3.27 for applications of this variant.)
Transcribed Image Text:3.18 Suppose o if r = 0, f(x) = t if ze ( . where n e N. Prove: f € R[0, 1], even though f has infinitely many discontinuities. (Hint: Consider Theorem 3.2.3.) For reference Theorem 3.2.3 If f is monotone on [a, b), then f € R[a, b). Proof: Suppose first that f is increasing. Then, for each partition P of (a, b] we have U(f, P) – L(f, P) =Ei (#+) – f(Ti-1)]A¤4 k=1 i=1 = ||P||S() – f(a)] → 0 as ||P|| → 0. Thus f € R(a, b). Finally if f is decreasing, then - f is integrable by the first part of this proof, and so f = -(-f) € R{a, b] as well. Here is a variant of the Darboux Integrability Criterion which is often useful. (See Exercises 3.26 and 3.27 for applications of this variant.)
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