Lecture 05 - Jan 14

Exercise:- If is a semi-algebra then

Proof. (The argument actually works for any collection .)

1. Because , monotonicity of the generator gives:

2. On the other hand, is a -algebra containing . In particular, it is an algebra containing , so by minimality of (the smallest algebra containing ), we have:

   Applying to both sides and using yields:

Combining the two inclusions gives the desired equality:

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Carathéodory extension theorem

Let be a nonempty set and be an algebra

Let be a measure i.e., is -additive such that

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def given from online Let be a nonempty set and be a semi-algebra.

Let be a set function that is -additive on . There exists a unique measure on the generated algebra such that:

For any , we can write as a finite disjoint union:

The measure on the algebra is then given by:

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Outer measure

An outer measure on is a function

satisfies

2. if
3. for every , (-subadditive)

Define

Lemma 1

is an outer measure.

Proof

Obviously,

Suppose then as the set over which the infimum is taken as the definition of includes the set corresponding set in the definition of

Suppose

Given , for each ,

Now,

and

Since (arbitrary),

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Define,

Remark:-

i) If , then

ii)

iii) To check it suffices to show that

Lemma 2

Proof

Take and with Given such that and Since and are covers of and respectively.

Lemma 3

is an algebra.

Proof

Take . Need to show:

Since ,

Since ,

NOTE

That,

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