Page 168 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 13: Signed Measures

Now, the set B – E is either negative or contains a subset of positive measure. If the set B – E is Notes
1 1
a negative set, then we may take A = B – E and the theorem is done. So, suppose that B – E is not
1 1
a negative set. Then there must exist a measurable subset E of B – E and a smallest positive
2 1
number n with a property
2
1
(E ) > .
2 n
2
Since B = (B – E E ) (E E ),
1 2 2 2
and (B – E E ) (E E ) = ,
1 2 1 2
we have (B) = (B – E E ) + (E E )
1 2 2 2
or (B – E E ) = (B) – (E E )
1 2 2 2
= (B) – (E ) – (E ).
1 2
As before, (B – E E ) > 0 [ (B) < 0, (E ) > 0 for r = 1, 2]
1 2 r
Thus, B – E E is a set of negative measure, which is either a negative set or contains a subset
1 2
of positive measure. If B – E E is a negative set, then the theorem is done by taking B = A – E
1 2 1
E . Otherwise we repeat the above process.
2
On repeating this process, at some stage we shall get either a negative subset A B s.t. (A) < 0
or a sequence <E > of disjoint measurable sets and a sequence <n : r N> of positive integers s.t.
r r
r 1 1
E B –  E n and < (E ) <
r n r
n 1 r
In first case, we have nothing to do. In the latter case, let

 
A = B – E n or B = A E n … (iii)
n 1 n 1
Then as before, it follows that

n .
(B) = (A) + (E )
n 1
1
> (A) + … (iv)
n
k 1 k
[ change of suffix is in material]
Since (B) is finite and assumes at most one of the values – and , it follows from (iv) that
1
(A) is finite and the series is convergent.
n
k 1 k

1
Then (A) < (B) –
n
k 1 k
= a finite negative number
( (B) is a finite negative number)

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