Page 252 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
Example: Show at the function f considered in example is not Riemann integrable.

b b b b
Solution: As shown in above example, f(x) dx = b a and f(x) dx 0 and so f(x) dx=  ¹  f(x) dx
-


a a a
and consequently f is not Riemann integrable.
h
Example: Show that a constant function is Riemann integrable over [a,b] and find f(x) dx.

1
b b
Solution: As proved in above example, f(x) dx k(b a)= - =  f(x) dx

a 1
h
Therefore, f is Riemann integrable on [a,b] and f(x) dx k(b a).= -

1
Theorem 1: If the partition P is a refinement of the partition P, of [a,b], then L(P ,f) £ L(P ,f) and
2 1 2
U(P ,f) £ U(P ,f).
2 1
Proof: Suppose P contains one point more than P,. Let this extra point be c. Let P = {x , x,, …, x }
2 1 0 n
and x < c < x . Let M and m be respectively the sup. f and inf. f in [x , x ]. Suppose sup. f and inf.
i-1 i i i i-1 i
f in [x , c] are p, and q and those in [c, x ] are p and q , respectively. Then,
i-1 1 i 2 2
-
-

-
+
L(P ,f) L(P ,f) = q (c x ) q (x - c) m x i
i
1
i
2
i 1
1
2
-
+
= (q - m )(c x ) (q - m )(x - c)
-
1 i i 1 2 i i
-
(since A x = (x - c) (c x - i 1 ) )
-
+
i
i
+
Similarly U(P ,f) U(P ,f) (p- = - M )(c x ) (p - M )(x - c)
-
2 1 1 i - i 1 2 i i
Now m £ q £ p £ M
i 1 1 i
m £ q £ p £ M i
i
2
2
Therefore
-

L(P ,f) L(P ,f) 0 and U(P ,f) U(P ,f) 0
£
-
2
1
1
2
Therefore

L(P ,f) L(P ,f) 0 and U(P ,f) U(P ,I).
-
-
1
2
1
2
Is P contains p points more than P , then adding these extra points one by one to P and using the
2 1 1
above results, the theorem is proved. We can also write the theorem as
£
£
L(P ,f) L(P ,f) U(P ,f) U(P ,f)
£
2
1
1
2
from which it follows that U(P ,f) L(P ,f) U(P ,f) L(P ,f).- 2 £ 1 - 1 As an illustration of theorem 1,
2
we consider the following example.
Example: Verify Theorem 1 for the function f(x) = x + 1 defined over [0, 1] and the
}
}
{
1 1 1 1 2 3
1 1 1 3
partition P 0, , , , ,1 and P = { 0, , , , , , ,1 .
1 4 3 2 4 2 6 4 3 2 3 4
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