Definition (Complete)
Let be a measure space. We say that it is complete if for all with and then .
Suppose is a measure space. Define
Note: is a subset of Null set where .
Lemma 1
is a -algebra and . {Prove}
Lemma 2
Define
Show that is a measure on and .
Theorem
is complete.
Proof Exercise.
Definition (Completion)
is called the completion of .
Note: , , which is a pre-measure.
i.e., if disjoint then
is a -algebra.
is a complete measure space.
Theorem
Suppose is a pre-measure, where is an algebra over . Then is the completion of .
Sketch of the Proof
i) Prove is complete.
ii) .
Remarks:- Suppose increasing, right continuous. Then we have a -algebra containing .
Moreover, is the completion of wrt .
Remarks
When , we define (denoted),
is called the Lebesgue measure -algebra,
and is called the Lebesgue measure.
Exercise
Lebesgue measure is translation invariant.
Note:-
i.e.
-
-
. Then
Exercise:
Any mapping between 2 sets & induces a mapping defined by
which preserves unions, intersection & complementary
i.e.
Remark
If is a -algebra over , then is also a -algebra.
Definition (Measurable)
If & are two measurable spaces, a mapping is called measurable, or just measurable if are understood if
Exercise
Composition of two measurable functions is measurable.
Note:- .
Proposition
If is generated by , then is measurable if
Proof:- Exercise.
Corollary
Suppose is continuous between , where are topological spaces. Show that is measurable wrt Borel -algebra or is -measurable.
Definition (Borel measurable)
A function is called measurable (or Borel measurable) if for all open sets in .
Hence,
Proposition
is measurable if and only if