Page 110 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 9: Measure Spaces

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Proof: Let  be a -algebra of subsets of the set Y. To show that f () is a -algebra of subsets of Notes
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the set D (f) we show that D (f) f (); if A f () then D (f)\A f (); and for any sequence
(A : n ) in f () we have U A f (B).
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n n  n
1. By (1) of above theorem, we have D (f) = f (Y) f (B) since Y .
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C
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C
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2. Let A f (). Then A = f () for some B . Since B  we have f (B ) f (). On the
C
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other hand by (2) of above theorem, we have f (B ) = D (f)\f (B) = D (f)\A. Thus D (f)\A
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f ().
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3. Let (A : n ) be a sequence in f (). Then A = f (B ) for some B  for each n .
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n n n n
Then by (3) of above theorem, we have
 A n  f (B ) f 1  B n f ( ) ,

1
1
n
n  n  n 


since B n ( ) .
n 
Measurable Mapping
Definition: Given two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and
 (f) Y. We say that f is a / measurable mapping if f (B) for every B , that is, f ()
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.
Theorem 4: Given two measurable spaces (X, ) and (Y, ). Let f be a /-measurable mapping.
(a) If , is a -algebra of subsets of X such that , , then f is  /-measurable.
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(b) If  is a -algebra of subsets of Y such that  , then f is / -measurable.
0 0 0
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Proof: (a) Follows from f ()   and (b) from f (B ) f () .
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Composition of two measurable mappings is a measurable mapping provided that the two
measurable mappings from a chain.
Theorem 5: Given two measurable spaces (X, ) and (Y, ), where = () and  is arbitrary
collection of subsets of Y. Let f be a mapping with D (f) and (f) Y. Then f is a /-
measurable mapping of D (f) into Y if and only if f () .
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Proof: If f is a /-measurable mapping of D (f) into Y, then f ()  so that f () .
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Conversely if f () , then (f () () = . Now by theorem,
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“Let f be a mapping of a set X into a set Y. Then for an arbitrary collection C of subsets of Y, we
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have (f ()) = f ( ().”
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(f () = f ( ()) = f (). Thus f ()  and f is a /- measurable mapping of D (f).
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Theorem 6: If X is a thick subset of a measure space (X, S, ), if S = S X , and if, for E in S, (E
0 0 0 0
X ) = (E), then (X , S , ) is a measure space.
0 0 0 0
Proof: If two sets, E and E , in S are such that E X = E X , then (E E ) X = 0, so that
1 2 1 0 2 0 1 2 o
(E E ) = 0 and therefore (E ) = (E ). In other words is indeed unambiguously defined on
1 2 1 2 0
S .
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Suppose next that {F } is a disjoint sequence of sets in S , and let E be a set in S such that
n 0 n
F = E X , n = 1, 2, … .
n n 0
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