Page 294 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Proof: Since f is integrable on [a,b], it is bounded. In other words, there exists a positive number
Î
M such that f(x) £ M, " x [a,b].
e
Let  > 0 be any number. Choose x,y Î [a,b], x < y, such that x y- < . Then
M
y x
F(y) F(x) = ò f(t) dt - ò f(t) dt
-
a a
x y x
= ò f(t) dt + ò f(t) dt - ò f(t) dt
a x a
y
= ò f(t) dt
x

y
£ ò f(t) dt
x
y
£ ò Mdt = M(y x) < Î
-
x
Similarly you can discuss the case when y < x. This shows that F is continuous on [a,b]. In fact this
proves the uniform continuity of F.
Now, suppose f is continuous at a point x of [a, b]
0
We can choose some suitable h  0 such that x + h Î [a, b].
0
Then,
0 x + h 0 x
F(x + h) – F(x ) = ò f(t) dt - ò f(t) dt
0 0
a a
0 x 0 x + h 0 x 0 x + h
= ò f(t) dt + ò f(t) d(t) - ò f(t) dt = ò f(t) dt
a 0 x a 0 x
Thus,

0 x + h
F(x + h) – F(x ) = ò f(t) dt …(1)
0 0
xn
Now

+
F(x + h) F(x ) 1 x0 h 1 0 x + h
-
0 0 - f(x ) = ò f(t) dt - ´ ò f(x ) dt
0
h 0 h x0 h 0
0 x + h
1
-
= ò [f(t) f(x )]dt .
0
h
0 x
Since f is continuous at x , given a number Î >0, 3 a number  > 0 such that f(x) f(x )- <  /2,
0 0
whenever x x- 0 <  and x Î [a,b]. So, if h <  then f(t) f(x ) < Î /2, for t Î [x ,x + h ], and
-
,
0
0
0
consequently
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