Page 352 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Notation: We shall denote the set of all (real-valued) bounded measurable functions defined on
E which vanishes outside a set of finite measure by B (E).
0
So from now on for f Î B (E), we have
0
A ò f = inf { A ò  : f   Î So (E) } = sup { ò  : f   Î So (E) }
A
for any A  E.
Note also that B (E) is a vector lattice, by which we mean it is a vector space partially ordered by
0
 (such that f  g if and only if f(x)  g(x) for all x Î E) and every two elements of it (say f, g Î B (E))
0
have a least upper bound in it (namely f V g Î B (E)). (Why is it a least upper bound?)
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We have the following nice proposition concerning the relationship between the Riemann and
the Lebsegue integrals.
Proposition: If f : [a, b]  is Riemann integrable on the closed and bounded interval [a, b], then
f Î B ([a, b]) and
0
b
(3) () f =  [a, b] f,
( )ò
a ò
where the () and () represents Riemann integral and Lebesgue integral respectively.
Proof: Since step functions defined on closed and bounded interval [a, b] are simple and have the
same Lebesgue and Riemann integral over [a, b] (why?), we see from the definitions
b b
() = ò f = sup { a ò  : f  step on [a, b] }
a
() = ò f = sup ò  : f   simple on [a, b] }
{ [a, b]
[a, b]
() = ò f = inf ò  : f   simple on [a, b] }
[a, b] { [a, b]
b b
() = ò f = inf { a ò  : f   step on [a, b] }
a
that

b       b
(4) () = ò f ( )ò f ( )ò [a, b] f ( )ò a f
a [a, b]
whenever the four quantities exist. Now if f is Riemann integrable over [a, b], then f is bounded
on [a,b]. Since [a,b] is of finite measure, we see that all four quantities in (4) exist. In that case

( )ò
( )ò b f =  b f as well so all four quantities in (4) are equal, which implies that f is measurable
a a
(so f Î B ([a, b])) and (3) holds.
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Proposition: Properties of the Lebesgue integral
Suppose f, g Î B (E). Then f + g, af, |f|Î B (E), and for any A  E, we have
0 0
(a) A ò (f g) = A ò f A ò + g
+
(b) A ò a = a A ò f for all a Î .
f
f
(c) A ò f = E ò  A
(d) If B  A then f + ò A\B f .
B ò
(e) If B  A and f  A ò f
B ò

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