Page 169 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes or (A) < 0.
Again, we know that difference of two measurable sets is measurable and enumerable union of
measurable sets is measurable therefore it follows from (iii) that A is a measurable set.

Now we shall prove that A is a negative set. Let E A be an arbitrary measurable set.


Since A = B – E n ,
n 1


E = B – E n .
n 1
Since n , we can choose k so large that
k
1
(E)
n
k 1
Letting n , we obtain
k
(E) 0.
Thus we have
(1) A is measurable.

(2) (E) 0, E A s.t. E is measurable.
Hence A is a negative set.

13.1.3 Hahn Decomposition Theorem

Theorem 3: Let be a signed measure on a measurable space (X, ). Then there exists a positive
set P and a negative set Q s.t.
P Q = and P Q = X.

Proof: Let (X, ) be a measurable space and let be a signed measure defined on a measurable
space (X, ). Since, by definition, assumes at most one of the values + or collection of all
negative subsets of X w.r.t. and let  be a collection of all negative subsets of X w.r.t. and let

k = inf { (E) : E )
(i) that there exists a sequence <E > in  such that
n
Lim (E ) = k.
n
n

Let Q = E n .
n 1
Since is a family of negative sets, < E > is a sequence of negative sets. Again, we know by
n
remark of theorem 1 that countable union of negative sets is negative, it follows that Q is a
negative subset of X so that
(Q) K.

162 LOVELY PROFESSIONAL UNIVERSITY