Page 167 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes C
where A (n 1,2,3 n 1) denotes complement of A (n = 1, 2, 3, … n – 1) with respect to X.
n n
Now, we know that complement of a measurable set is also measurable so that each
C
A (n 1,2,3 n 1) is measurable relative to . Again, intersection of countable collection of
n
measurable sets is also measurable. Hence B is a measurable subset of the positive set A . Thus
n n
(B ) 0 (by the definition of positive set) … (i)
n
Obviously, the set B are disjoint and
n

if B = B n , we get … (ii)
n 1

(B) = (B ) … (iii)
n
n 1
In view of (i).
(B) 0.
Thus, we have
(1) A is measurable for

A is a positive set A is a measurable set
n n

countable union A is measurable,
n
n 1
A =  A n is measurable
n 1

(2) (B) 0, B A s.t. B is a measurable set.
Hence A is a positive set, by definition.
Theorem 2: Let (X, ) be a measurable space and let be a signed measure defined on (X, A). If B
is a measurable set with finite negative measure i.e., – < (B) < 0, then prove that B contains a
negative set A B with the property (A) < 0.
Proof: If B is itself a negative set, then we may take A = B and theorem is done. Therefore
consider the case when B is not a negative set. Then there must exist a measurable subset E B
1
and a smallest positive integer n , s.t.
1
1
(E ) >
1 n
1
 B = (B – E ) E and (B – E ) E = ,
1 1 1 1
(B) = (B – E ) + (E ) … (i)
1 1
or (B – E ) = (B) – (E ) … (ii)
1 1
Since (B) is finite, (i) implies that (B – E ) and (E ) are finite. Again (B) < 0, (ii) implies that
1 1
(B – E ) < 0.
1

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