Page 257 - DMTH401_REAL ANALYSIS

Unit 20: The Riemann Integration

Notes
Example: Show that the function f where f(x) = [x] is integrable in [0,2] where [x] denotes
the greatest integer not greater than x.

0 if 0 £ x < 1
Solution: [x] = 1 if 1 £ x < 2
2 if x = 2

The points of discontinuity of f in [0,2] are 1 and 2 which are finite in number and so it is
integrable in [0,2].

Example: Show that the function F defined on the interval [0,1] by
ì 1 1
ï 2rx, when < x £ , where r is a positive integer
+
F(x) = í r 1 r
ï 0, elsewhere,
î
is Riemann integrable.

1 1
.
Solution: The function F is discontinuous at the points 0, 1, , , ¼ The set of points of
2 3
discontinuity has 0 as the only limit point. So, the limit points are finite in number and hence the
function F is integrable in [0,1], by Theorem 8.

There is one more class of integrable functions and this class is that monotonic functions. This
we prove in the following theorem.
Theorem 9: Every monotonic function is integrable.
Proof: We shall prove the theorem for the case where I: [a,b]  R is a monotonically increasing
function. The function is bounded. f(a) and f(b) being g.l.b. and l.u.b. Let Î > 0 be given number,
Let n be a positive integer such that

(b a)[f(b) f(a)]
-
-
n >
Î
Divide the interval [a,b] into n equal sub-intervals, by the partition P = {x , x …, x } of [a, b]. Then
0 1 0
n
-
U(P,f) L(P,f) = å (M - m )( x )

i
i
i
i 1
=
-
b a n
-
= å [f(xi) f(xi 1)]
-
n i
(b a)
-
= [f(b) f(a)] <Î .
-
n
This proves that f is integrable. Discuss the case of monotonically decreasing function as an
exercise. Do it by yourself.
Exercise: Show that a monotonically decreasing function is integrable.
Now we give example to illustrate the theorem.
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