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Let F be a s-ring in X. Let f : X -> |R*. Let M be dense in |R.

(1) /\a:-M {x:-X : f(x) > a} :- F

What can we conclude?

(1) /\a:-M {x:-X : f(x) > a} :- F

What can we conclude?

(2) /\a:-|R {x:-X : f(x) > a} :- F

page 65 in 2nd measure

page 65 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Prove that y is additive on M.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Prove that y is additive on M.

Let A:-M and AnB=O.

Then y(AuB) = y( (AuB)nA ) + y( (AuB)\A ).

Notice that y( (AuB)nA ) = y(A).

Since AnB=O, we get y( (AuB)\A ) = y(B).

Hence y(AuB) = y(A) + y(B). This is enough to prove additivity.

page 41 in 2nd measure

Then y(AuB) = y( (AuB)nA ) + y( (AuB)\A ).

Notice that y( (AuB)nA ) = y(A).

Since AnB=O, we get y( (AuB)\A ) = y(B).

Hence y(AuB) = y(A) + y(B). This is enough to prove additivity.

page 41 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E) + y(A n F)

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E) + y(A n F)

/\AcX y( A n (E u F) ) = y(A n E) + y(A n F)

Take any AcX.

Since E:-M, we have y(An(EuF)) = y(An(EuF) n E) + y(An(EuF) \ E).

Since EnF=O, we get y(An(EuF)) = y(A n E) + y(A n F).

page 42 in 2nd measure

Take any AcX.

Since E:-M, we have y(An(EuF)) = y(An(EuF) n E) + y(An(EuF) \ E).

Since EnF=O, we get y(An(EuF)) = y(A n E) + y(A n F).

page 42 in 2nd measure

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E n F) + y(A n E n F') + y(A n E' n F)

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let E,F belong to M and E n F = O.

Prove that:

/\AcX y( ??? ) = y(A n E n F) + y(A n E n F') + y(A n E' n F)

/\AcX y(A n (E u F) ) = y(A n E n F) + y(A n E n F') + y(A n E'
n F')

page 40 in 2nd measure

the line following two stars: (**)

page 40 in 2nd measure

the line following two stars: (**)

Let X be a set. Let y : P(X)->[0,oo] , y(O)=0.

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[1],E[2],...,E[n] } be a finite disjoint collection of sets in M.

Prove that:

/\AcX y( ??? ) = +(k=1 to k=n) y(A n E[k])

M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[1],E[2],...,E[n] } be a finite disjoint collection of sets in M.

Prove that:

/\AcX y( ??? ) = +(k=1 to k=n) y(A n E[k])

/\AcX y( A n E ) = +(k=1 to k=n) y(A n E[k])

where E = E[1] u E[2] u ... u E[n]

page 43 in 2nd measure

where E = E[1] u E[2] u ... u E[n]

page 43 in 2nd measure

Let y : P(X) -> |R* such that

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

page 44 in 2nd measure

Let y : P(X) -> |R* such that

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[n] }(n:-|N) be a disjoint collection of sets in M.

Let E be the union of this collection.

(1) y(O) = 0

(2) if A c B c X, then y(A)<=y(B)

(3) y is countably subadditive

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

Let { E[n] }(n:-|N) be a disjoint collection of sets in M.

Let E be the union of this collection.

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n])
+ ...

Conclude from this that y is countably additive on M.

page 94, 47 in 2nd measure

Conclude from this that y is countably additive on M.

page 94, 47 in 2nd measure

Let y:P(X)->|R* be such that

(2) if A c B c X, then y(A)<=y(B)

(3) y is subadditive

Let E be a subset of X such that y(E) = 0.

What can we conclude?

(2) if A c B c X, then y(A)<=y(B)

(3) y is subadditive

Let E be a subset of X such that y(E) = 0.

What can we conclude?

/\AcX y(A) = y(AnE) + y(A\E)

PROOF

A = (A n E) u (A \ E)

y(A) <= y(A n E) + y(A \ E) <= y(E) + y(A) = y(A).

We showed y(A) = y(A n E) + y(A \ E).

page 49 in 2nd measure

PROOF

A = (A n E) u (A \ E)

y(A) <= y(A n E) + y(A \ E) <= y(E) + y(A) = y(A).

We showed y(A) = y(A n E) + y(A \ E).

page 49 in 2nd measure

Let y be an outer measure on X.

Define the collection of all measurable subsets of X.

State without proof four important theorems about this collection.

Define the collection of all measurable subsets of X.

State without proof four important theorems about this collection.

Let M = { EcX : /\AcX y(A) = y(AnE) + y(A\E) }.

By definition, E is measurable <=> E:-M.

(1) M is a s-algebra

(2) If { E[n] : n:-|N } is a disjoint collection of sets in M,

and if E is the union of this collection, then

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n]) + ...

By definition, E is measurable <=> E:-M.

(1) M is a s-algebra

(2) If { E[n] : n:-|N } is a disjoint collection of sets in M,

and if E is the union of this collection, then

/\AcX y(A n E) = y(A n E[1]) + y(A n E[2]) + ... + y(A n E[n]) + ...

+(n=1 to n=oo) (-1)^(n+1) / n = ???

+(n=1 to n=oo) (-1)^(n+1) / n = log(2)

page 83 in OLDTIMER

page 83 in OLDTIMER