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Let f:|R->|R be continuous.

Does there exist a diffable function F:|R->|R

such that /\x:-|R F'(x) = f(x) ?

Does there exist a diffable function F:|R->|R

such that /\x:-|R F'(x) = f(x) ?

YES.

F(x) = Integral( 0, x, f(t), dt )

page 92 in OLDTIMER

F(x) = Integral( 0, x, f(t), dt )

page 92 in OLDTIMER

Let f : |R -> |R be continuous. Let a:-|R.

Let g : |R -> |R be defined g(x) = Integral( a, x, f(t), dt ).

What can we conclude from this?

Let g : |R -> |R be defined g(x) = Integral( a, x, f(t), dt ).

What can we conclude from this?

/\x:-|R g'(x) = f(x)

page 92 in OLDTIMER

page 92 in OLDTIMER

Let a,b:-|R and a<b.

Let f : ]a,b[ -> |R be continuous and bounded.

Does there exist a diffable function g : ]a,b[ -> |R

such that g'(x) = f(x) for a<x<b ?

Let f : ]a,b[ -> |R be continuous and bounded.

Does there exist a diffable function g : ]a,b[ -> |R

such that g'(x) = f(x) for a<x<b ?

YES

page 92 in OLDTIMER

page 92 in OLDTIMER

(1) a,b:-|R, a!=b

(2) A,B :- |R, A<B

(3) g : *[a,b]* -> |R continuously diffable

(4) f : [A,B] -> continuous

(5) g(*[a,b]*) c [A,B]

What can we conclude from this?

(2) A,B :- |R, A<B

(3) g : *[a,b]* -> |R continuously diffable

(4) f : [A,B] -> continuous

(5) g(*[a,b]*) c [A,B]

What can we conclude from this?

Integral( g(a), g(b), f(x), dx ) = Integral( a, b,
f(g(x))*g'(x), dx )

page 93 in OLDTIMER

page 93 in OLDTIMER

(1) a,b :- |R, a < b

(2) g : [a,b] -> |R continuously diffable

(3) f : |R -> |R continuous

(4) g(a) = g(b)

Integral( a, b, f(g(x))*g'(x), dx ) = ???

(2) g : [a,b] -> |R continuously diffable

(3) f : |R -> |R continuous

(4) g(a) = g(b)

Integral( a, b, f(g(x))*g'(x), dx ) = ???

Integral( a, b, f(g(x))*g'(x), dx ) = 0

page 94 in OLDTIMER

page 94 in OLDTIMER

Let a[1], a[2], a[3], ..., a[n], a[n+1], ..., a[2n] be numbers.

+(k=1 to k=2n) a[k] = +(k=1 to k=n) ???

+(k=1 to k=2n) a[k] = +(k=1 to k=n) ???

+(k=1 to k=2n) a[k] = +(k=1 to k=n) a[2k-1] + a[2k]

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/F:-W F is closed and ??????????????

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/F:-W F is closed and ??????????????

/\A:-W /\E>0 \/F:-W F is closed and FcA and m(A) <= m(F) + E.

page 106 in 1st measure

page 106 in 1st measure

Consider the p-dimensional Euclidean space |R^p.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/H:-W ???????????? and m(A) <= m(H) + E.

Let W be the collection of all intervals in |R^p.

In the context of measure theory, let m be the "volume" function on W.

Complete this important theorem:

/\A:-W /\E>0 \/H:-W ???????????? and m(A) <= m(H) + E.

/\A:-W /\E>0 \/H:-W H is closed and HcA and m(A) <= m(H) + E.

page 106 in 1st measure

page 106 in 1st measure

Let K be a s-algebra in X. Let f,g : X -> |R* be K-measurable.

Can we conclude that function max(f,g) is K-measurable?

Can we conclude that function max(f,g) is K-measurable?

YES: max(f,g) is K-measurable

Take any a:-|R. Since f,g are K-measurable, {x : f(x)>a}:-K and {x : g(x)>a}:-K. Since K is an algebra, {x : f(x)>a or g(x)>a}:- K. Hence {x : max(f(x),g(x))>a}:-K. We showed that max(f,g) is K-measurable.

Take any a:-|R. Since f,g are K-measurable, {x : f(x)>a}:-K and {x : g(x)>a}:-K. Since K is an algebra, {x : f(x)>a or g(x)>a}:- K. Hence {x : max(f(x),g(x))>a}:-K. We showed that max(f,g) is K-measurable.

Let K be a s-algebra in X. Let f,g : X -> |R* be K-measurable.

Can we conclude that function min(f,g) is K-measurable?

Can we conclude that function min(f,g) is K-measurable?

YES: min(f,g) is K-measurable

Take any a:-|R. Since f,g are K-measurable, {x : f(x)<a}:-K and {x : g(x)<a}:-K. Since K is an algebra, {x : f(x)<a or g(x)<a}:- K. Hence {x : min(f(x),g(x))<a}:-K. We showed that min(f,g) is K-measurable.

Take any a:-|R. Since f,g are K-measurable, {x : f(x)<a}:-K and {x : g(x)<a}:-K. Since K is an algebra, {x : f(x)<a or g(x)<a}:- K. Hence {x : min(f(x),g(x))<a}:-K. We showed that min(f,g) is K-measurable.