Page 349 - DMTH401_REAL ANALYSIS

Unit 29: The Lebesgue Integral of Bounded Functions

Notes
m
A ò  = å b m(B Ç A) (by definition of the integral)
j
j
j 1
=
m æ ö
= å b m ç  E Ç A ÷ (by (1))
j
i
j 1 è {i:a i b }= j ø
=
m
= å b j å m(E Ç A) (by finite additivity of m)
i
j 1 {i:a i b }= j
=
m
= å b j å a m(E Ç A)
i
i
j 1 {i:a i b }= j
=
n
= å a m(E Ç A) (by (2))
i
i
=
i 1
This complete our proof.
29.2 Properties of the Lebesgue Integral
Proposition: (Properties of the Lebesgue integral)
Suppose  +  Î S (E). Then for any A E,
0
)
(a) A ò ( +  = A ò  + A ò  (Note that  +  ÎS (E) too be the vector space structure
0
(b) A ò a = a  for all a Î . (Note aÎS (E) again.)
A ò
0

(c) If a  a.e. on A then ò   A ò  .
A
(d) If  =  a.e. on A then ò  = A ò  .
A
0
(e) If  0 a.e. on A and ò  = , then  = 0 a.e. on A.
A
(f) A ò   A ò | |.(Note| | So (E) too.Why?)
Î


Remark: (a) and (b) are known as the linearity property of the integral, while (c) is known as the
monotonicity property. Furthermore, Lemma is now seen to hold by the linearity of the integral
even without the disjointness assumption on the E ’s.
i
Proof:
(a) Let  = å n i 1 a  and  = å m j 1 b  be canonical representations of  and  respectively.
j
Bj
=
=
i
Ai
j 1 
Then noting that  A i = å m = i A Ç for all i and  = å n i 1  A i Ç j B
j B
=
j B
n n m
 = å a  A = å å a  A i Ç j B
i
i
i 1 i i 1 j 1
=
=
=
m n m
 = å b  = å å b  A i BÇ
j
j
j B
=
=
j 1 i 1 j 1 j
=
Consequently
n m
 +  = å å (a + b ) A i Ç j B .
j
i
=
=
i 1 j 1
LOVELY PROFESSIONAL UNIVERSITY 343