Page 348 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Proof: With the above proposition we see that every simple function  can be written uniquely in
the form

n
 = å a  i E
i
=
i 1
–1
where the a ’s are all non-zero and distinct, and the E ’s are disjoint. (Simply take E ’s  {a } for
i i i i
i = 1, 2,...., n where a , a ,...., a are all the distinct values of .) We say this is the canonical
1 2 n
representation of .
We adopt the following notation:
Notation: A function f : E   is said to vanish outside a set of finite measure if there exists a set
A with m(A) <  such that f vanishes outside A, i.e.
f = 0 on E\A
or equivalently f(x) = 0 for all x Î E\A. We denote the set of all simple functions defined on E
which vanish outside a set of finite measure by S (E). Note that it forms a vector space.
0
We are now ready for the definition of the Lebesgue integral of such functions.
Definition: For any Î S (E) and any A  E, we define the Lebesgue integral of  over A by
0
n
A ò  = å a m(E Ç A)
i
i
=
i 1
n
where  = å a  i E is the canonical representation of . (From now on we shall adopt the
i
i 1
=
convention that 0   = 0. We need this convention here because it may happen that one a is 0
i
while the corresponding E C\A has infinite measure. Also note that here A is implicitly assumed
i
to be measurable so m(E n A) makes sense. We shall never integrate over non-measurable sets.)
i
It follows readily from the above definition that
A ò  = A ò  A
for any Î S (E) and for any A  E.
0
We now establish some major properties of this integral (with monotonicity and linearity being
probably the most important ones). We begin with the following lemma.
Lemma: Suppose  = å n i 1 a  Î S (E) where the E ’s are disjoint. Then for any A  E,
=
Ei
i
i
0
A ò  = å n i 1 a m(E Ç A)
=
i
i
holds even if the a ’s are not necessarily distinct.
i
Proof: If  = å n j 1 b  is the canonical representation of , we have
Bj
i
=
m
1. B =  E i
j {i :a i b }= j
for j = 1, 2,..., m and
m
2. {1, 2,..., n} =  {i : a = b }
j ,
i
j 1
=
where both unions are disjoint unions. Hence for any A  E, we have
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