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Lecture 04 - Jan 13

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Theorem

Let be a measure space. Then

  1. (monotonically). If with then
  2. (subadditivity). then
  3. (continuity from below). If with that is and Then
  4. (Continuity from above):- If with . That is, and and . Then

Example:- . Consider Then But and .

Proof

ii) Let , .

Then ‘s are disjoint and .

Therefore:

iii) Set . Then is a disjoint collection in with

iv) Set ; Then is an increasing sequence to .

Theorem

Let be a semi algebra and be an additive (respectively -additive) set function. Then there exists a set function which is additive (respectively -additive):

  2. If there exist two additive (respectively -additive) set functions and on satisfying (1) then Proof

If then for , Lecture 3

We define,

Suppose where ,

NOTE

To prove

Note that is a disjoint union of elements from .

Therefore, // Interchanging the role of and //

Take such that

Let,

with . Then we can write

Then,

is a disjoint union of sets.

ii) Suppose

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