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The Riemann Integral

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14 1. <strong>The</strong> <strong>Riemann</strong> <strong>Integral</strong><br />

1.2<br />

1<br />

0.8<br />

y<br />

0.6<br />

0.4<br />

0.2<br />

0<br />

0 0.2 0.4 0.6 0.8 1<br />

x<br />

Figure 2. <strong>The</strong> graph of the monotonic function in Example 1.22 with a countably<br />

infinite, dense set of jump discontinuities.<br />

<strong>The</strong> proof for a monotonic decreasing function f is similar, with<br />

sup<br />

I k<br />

f = f(x k−1 ),<br />

inf<br />

I k<br />

f = f(x k ),<br />

or we can apply the result for increasing functions to −f and use <strong>The</strong>orem 1.23<br />

below.<br />

□<br />

Monotonic functions needn’t be continuous, and they may be discontinuous at<br />

a countably infinite number of points.<br />

Example 1.22. Let {q k : k ∈ N} be an enumeration of the rational numbers in<br />

[0,1) and let (a k ) be a sequence of strictly positive real numbers such that<br />

∞∑<br />

a k = 1.<br />

Define f : [0,1] → R by<br />

f(x) = ∑<br />

k∈Q(x)<br />

a k ,<br />

k=1<br />

Q(x) = {k ∈ N : q k ∈ [0,x)}.<br />

for x > 0, and f(0) = 0. That is, f(x) is obtained by summing the terms in the<br />

series whose indices k correspond to the rational numbers 0 ≤ q k < x.<br />

For x = 1, this sum includes all the terms in the series, so f(1) = 1. For<br />

every 0 < x < 1, there are infinitely many terms in the sum, since the rationals<br />

are dense in [0,x), and f is increasing, since the number of terms increases with x.<br />

By <strong>The</strong>orem 1.21, f is <strong>Riemann</strong> integrable on [0,1]. Although f is integrable, it<br />

has a countably infinite number of jump discontinuities at every rational number<br />

in [0,1), which are dense in [0,1], <strong>The</strong> function is continuous elsewhere (the proof<br />

is left as an exercise).

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