Page 259 - DMTH401_REAL ANALYSIS

Unit 20: The Riemann Integration

Notes
b b
   f(x) dx =   f(x) dx
a a
Similarly L(P, f) =  U L(P,f).

h b

   f(x) dx = A f(x) dx
a a
Since f is integrable in [a,b], therefore
6 b b
 f(x) dx =  f(x) dx =  f(x) dx.
a a a

6 6 b
Hence   f(x) dx =  f(x) dx =   f(x) dx,

a
whether  0 or  < 0.
b b
f R [a,b]and  
Hence  Î f(x) dx =   f(x) dx.
a a
Now suppose that  = –1. In this case the theorem says that if f Î R [a,b], then ( f) R[a,b]- Î

b b
-
 [ f(x)]dx =  f(x) dx.
a - a
Theorem 11: If f Î R [a,b], g Î R [a,b],. then f + g Î R [a,b] and
b b b
+
 (f g)(x)dx  = f(x) dx +  g(x) dx.
a a
Proof: We first show that f+g Î R [a,b]. Let Î > 0 be a given number. Since f Î R [a,b], g Î R [a,b],
there exist partitions P and Q of [a,b] such that U(P,f) – L(Pf) < Î/2 and U (Q,g) – L (Q,g)
< Î/2
If T is a partition of [a,b] which refines both P and Q, then
U(T,f) – L(T,f) < Î/2 [U(T,f) – L(T,f) £ U(P,f) – L(P,f)].
Similarly,

U(T,g) – L(T,g) < Î/ 2 (5)
Also note that, if M = sup {f(x) : x £ x £ x }
i i-1 i
and
N = sup {g(x): x 6 x £ x }
i i-1 i
then,
sup {f(x) + g(x): x £ x £ x } £ M + N .
i-1 i 1 i
Using this, it readily follows that
U(T, f+g) £ U(T,f) + U(T,g)
for every partition T of [a,b]. Similarly

L(T,f+g) L(T,f) + L(T,g)
for every partition T of [a,b].
Thus U (T,f+g) – L (T,f+g) £ [U(T,f) + U(T,g) – L [(T,f) + L(T,g)]

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