Page 258 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
1
Example: Show that the function f defined by the condition f(x) =
2 n
1 1
when < x £ ,n = 0,1,2 ¼
+
2 n 1 2 n
is integrable in [0,1]

Solution: Here we have f (0) = 0,
1
f(x) = 1 when < x £ 1
2
2
1 æ 1 ö 1
f(x) = when ç ÷ < x £
2 è 2 ø 2
------------------------------------------------
------------------------------------------------
Clearly f is monotonically increasing in [0, 1]. Hence it is integrable.

20.3 Algebra of Integrable Functions

As we discussed the algebra of the derivable functions. Likewise, we shall now study the algebra
of the integrable functions. In the previous class, you have seen that there are integrable as well
as non-integrable functions. In this section you will see that the set of all integrable functions on
[a,b] is closed under addition and multiplication by real numbers, and that the integral of a sum
equals the sum of the integrals. You will also see that difference, product and quotient of two
integrable functions is also integrable.
All these results are given in the following theorems.
Theorem 10: If f Î R (a , b), and  is any real number, then  f Î R (a,b) and

b b
  f(x) dx =   f(x) dx.
a a
Proof: Let P = {x,, x ,...., x ) be a partition of [a,b]. Let M and m be the respective l.u.b. and g.l.b. of
1 n i i
the function f in [x , x ]. Then  M and  m are the respective l.u.b. and g.l.b. of the function  f
i-1 i i i
in [x , x ], if A  0, and  m and  M are the respective l.u.b. and g.l.b. of h f in [x , x ], if
i-1 i i i i-1 i
h < 0.
n n
When   0, then U(P, f) = å A M  x =  å M Ax, =  U(P,f).

i
i
i
i 1 i 1
=
=
6 b
   f(x) dx =   f(x) dx.
a
Similarly L(P,  f) = A L(P,f).
b b
   f(x) dx =   f(x) dx
a a
n
If  < 0, U(P, f) = å  m  x =  L(P,f).

i
i
i 1
=
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