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Let F be a s-algebra in X.

Let f : X -> |R \ {-oo,oo} be F-measurable.

Prove that function f*f is F-measurable.

Let f : X -> |R \ {-oo,oo} be F-measurable.

Prove that function f*f is F-measurable.

Take any a:-|R. If a<0, then {x: f^2(x) >= a} :- F.

If a>=0, then {x: f^2(x) > a} = {x: |f(x)| > sqrt(a)} =

= {x: f(x) > sqrt(a) or f(x) < -sqrt(a)} =

= {x: f(x) > sqrt(a)} u {x: f(x) < -sqrt(a)} :- F

If a>=0, then {x: f^2(x) > a} = {x: |f(x)| > sqrt(a)} =

= {x: f(x) > sqrt(a) or f(x) < -sqrt(a)} =

= {x: f(x) > sqrt(a)} u {x: f(x) < -sqrt(a)} :- F

Let F be a s-algebra in X.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

Prove that f*g is F-measurable by using a trick.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

Prove that f*g is F-measurable by using a trick.

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

Use the facts that f+g, f-g are F-measurable,

and that if h:X->|R is measurable then so is h^2.

page 138 in 1st measure

Use the facts that f+g, f-g are F-measurable,

and that if h:X->|R is measurable then so is h^2.

page 138 in 1st measure

Let F be a s-algebra in X.

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are the two ways of proving that f*g is F-measurable?

Let f,g : X -> |R \ {-oo,oo} be F-measurable.

What are the two ways of proving that f*g is F-measurable?

WAY 1:

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x*y (page 138 in 1st measure)

WAY 2:

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

(1) F : |R^2 -> |R is continuous

(2) h : X -> |R , h(x) = F(f(x),g(x))

thesis: h is F-measurable

Let F(x,y)=x*y (page 138 in 1st measure)

WAY 2:

f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]

Let W c P(X). Let U c X.

Suppose that /\(a:-U) \/(B:-W) a:-B and B c U.

Can we conclude that U is W-open?

Suppose that /\(a:-U) \/(B:-W) a:-B and B c U.

Can we conclude that U is W-open?

YES.

page 78,79 in gen top

page 78,79 in gen top

f : [0,oo[ -> |R , f(x) = 1/(1+sqrt(x))

Find a diffable function F : [0,oo[ -> |R

such that /\x>=0 F'(x) = f(x).

Find a diffable function F : [0,oo[ -> |R

such that /\x>=0 F'(x) = f(x).

F(x) = 2 * ( sqrt(x) - log(1+sqrt(x) )

page 116 in OLDTIMER

page 116 in OLDTIMER

(1) 0 < a <= oo

(2) F,f : [0,a) -> |R

(3) /\0<x<a F'(x) = f(x)

(4) F and f are both continuous at 0

(5) F(0) = 0

What can we conclude from this?

(2) F,f : [0,a) -> |R

(3) /\0<x<a F'(x) = f(x)

(4) F and f are both continuous at 0

(5) F(0) = 0

What can we conclude from this?

F'(0) = f(0)

page 117 in OLDTIMER

page 117 in OLDTIMER

Prove that |R is homeomorphic with the bounded open interval (-
1;1).

Do not use trigonometric functions.

Do not use trigonometric functions.

f : |R -> (-1;1) , f(x) = x / ( |x|+1 )

f is 1-1, onto, continuously diffable

f is 1-1, onto, continuously diffable

Let H be a s-algebra in X.

Let f : X -> |R \ {-oo,oo} be H-measurable.

Let F : |R -> |R be a Borel function.

Let g : X -> |R , g(x) = F(f(x)).

What can we conclude?

Let f : X -> |R \ {-oo,oo} be H-measurable.

Let F : |R -> |R be a Borel function.

Let g : X -> |R , g(x) = F(f(x)).

What can we conclude?

function g is H-measurable

Take any a:-|R. Let G={x:-|R : F(x)>a}. G is Borel in |R.

Function f is H-measurable, hence f-1(G) :- H.

{x:-X : g(x)>a} = {x:-X : F(f(x))>a} = {x:-X : f(x) :- G} = f- 1(G) :- H

Take any a:-|R. Let G={x:-|R : F(x)>a}. G is Borel in |R.

Function f is H-measurable, hence f-1(G) :- H.

{x:-X : g(x)>a} = {x:-X : F(f(x))>a} = {x:-X : f(x) :- G} = f- 1(G) :- H

In the context of measure theory, what does it mean that a
function X->|R* is simple?

(0) f : X -> |R* is simple

iff

(1) /\x:-X f(x):-|R

(2) f(X) is finite

===================

f(X) = { a[1], a[2], ..., a[n] } c |R

iff

(1) /\x:-X f(x):-|R

(2) f(X) is finite

===================

f(X) = { a[1], a[2], ..., a[n] } c |R

(1) f : X -> |R*.

(2) n:-|N

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

(4) A[1], ..., A[n] c X , none of them is empty

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

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

(2) n:-|N

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

(4) A[1], ..., A[n] c X , none of them is empty

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

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

f(X) = { a[1] , ... , a[n] } c |R

f is a simple function

page 105 in 2nd measure

page 140 in 1st measure

f is a simple function

page 105 in 2nd measure

page 140 in 1st measure