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Let F be a s-algebra in X.

(1) f : X -> |R*

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(1) f : X -> |R*

(2) n:-|N

(3) a[1], ..., a[n] :- |R

(4) A[1], ..., A[n] :- F

(5) X = \\*//(k=1 to n) A[k]

(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

f is simple and F-measurable

page 106 in 2nd measure

page 141 in 1st measure

page 106 in 2nd measure

page 141 in 1st measure

Let F be a s-algebra in X.

Suppose that f : X -> |R is simple and F-measurable.

How can function f be conveniently expressed?

Suppose that f : X -> |R is simple and F-measurable.

How can function f be conveniently expressed?

There exist:

(1) n:-|N

(2) a[1], ... , a[n] :- |R ; all different ; n numbers altogether

(3) A[1], ... , A[n] :- F ; disjoint and non-empty

such that

(4) X = \\*//(k=1 to n) A[k]

(5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

(1) n:-|N

(2) a[1], ... , a[n] :- |R ; all different ; n numbers altogether

(3) A[1], ... , A[n] :- F ; disjoint and non-empty

such that

(4) X = \\*//(k=1 to n) A[k]

(5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that A[n] is increasing?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that A[n] is increasing?

YES

page 99 in 2nd measure

page 99 in 2nd measure

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

What can we conclude from this?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

What can we conclude from this?

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

page 99 in 2nd measure

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

page 99 in 2nd measure

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that { A[k]\A[k-1] }(k:-|N) is a disjoint sequence?

/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.

Can we conclude that { A[k]\A[k-1] }(k:-|N) is a disjoint sequence?

YES. page 99 in 2nd measure

We can even conclude much more:

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

We can even conclude much more:

(1) /\n:-|N A[n] c A[n+1]

(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence

(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O

(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

{A[k]\A[k-1]}(k:-|N) is a disjoint sequence.

Can we conclude that A[n] is increasing?

{A[k]\A[k-1]}(k:-|N) is a disjoint sequence.

Can we conclude that A[n] is increasing?

NO

A[n] = {n}

A[n] = {n}

Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such
that

\\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Can we conclude that A[n] is increasing?

\\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.

Can we conclude that A[n] is increasing?

NO

A[n] = {n}

A[n] = {n}

If a :- [-oo,oo], what does (a+) mean?

(a+) = max(0 , a)

page 103 in 2nd measure

page 103 in 2nd measure

If a :- [-oo,oo], what does (a-) mean?

(a-) = max(0 , -a)

page 103

page 103

In the context of measure theory,

what does it mean: f~g ?

what does it mean: f~g ?

In this context:

(1) F is a s-algebra in X

(2) y : F -> [0,oo] is countably additive

(3) f,g : X -> |R* are F-measurable

f~g <=> y( {x:-X : f(x)!=g(x)} ) = 0

(1) F is a s-algebra in X

(2) y : F -> [0,oo] is countably additive

(3) f,g : X -> |R* are F-measurable

f~g <=> y( {x:-X : f(x)!=g(x)} ) = 0