The Riemann Integral
The Riemann Integral
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).