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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 , some of them may be empty

(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 , some of them may be 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 is simple and F-measurable

page 109 in 2nd measure

page 109 in 2nd measure

Let F be a s-algebra in X.

Let f : X -> [0,oo] be F-measurable.

In this setting, how can we interestingly approximate function f ?

(In this item, prove the existence of this interesting approximation.)

Let f : X -> [0,oo] be F-measurable.

In this setting, how can we interestingly approximate function f ?

(In this item, prove the existence of this interesting approximation.)

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple and measurable

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 113 in 2nd measure

(1) /\n:-|N s[n] : X -> [0,oo) is simple and measurable

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 113 in 2nd measure

f : X -> [0,oo]

In this broad setting, how can we interestingly approximate function f ?

(In this item, do not prove the existence of this interesting approximation.)

In this broad setting, how can we interestingly approximate function f ?

(In this item, do not prove the existence of this interesting approximation.)

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 110 in 2nd measure

(1) /\n:-|N s[n] : X -> [0,oo) is simple

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

page 110 in 2nd measure

Let f : X -> [0,M] , M < oo.

In this broad setting, how can we interestingly approximate function f ?

In this broad setting, how can we interestingly approximate function f ?

There exists a sequence of functions { s[n] }(n:-|N) such that

(1) /\n:-|N s[n] : X -> [0,oo) is simple and bounded

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

(4) sequence s[n] converges uniformly to function f

page 113 in 2nd measure

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

(1) /\n:-|N s[n] : X -> [0,oo) is simple and bounded

(2) /\x:-X /\n:-|N s[n](x) <= s[n+1](x)

(3) /\x:-X lim(n->oo) s[n](x) = f(x)

(4) sequence s[n] converges uniformly to function f

page 113 in 2nd measure

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

lim(x,y)->(0,0) y*(x-sin(x)) / (x^4 + y^2) = ???

lim(x,y)->(0,0) y*(x-sin(x)) / (x^4 + y^2) = 0

hint:

x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6

|xy/(x^2+y^2)|<=1/2

page 100 in OLDTIMER

hint:

x-sin(x) = x^3 * u(x), where lim(x->0) u(x) = 1/6

|xy/(x^2+y^2)|<=1/2

page 100 in OLDTIMER

Let a : |N -> |C.

Let z be a complex number.

Suppose that series a[n]*|z|^n converges.

Can we conclude that series a[n]*z^n converges?

Let z be a complex number.

Suppose that series a[n]*|z|^n converges.

Can we conclude that series a[n]*z^n converges?

NO

(1) Let a[n] = 1/n * (-1)^n. Let z = (-1).

(2) Consider series (-1)^n / n * z^(3n). Let z = -1.

(1) Let a[n] = 1/n * (-1)^n. Let z = (-1).

(2) Consider series (-1)^n / n * z^(3n). Let z = -1.

(1) W is a ring of sets in X

(2) y : W -> [0,oo) is additive

(3) A[1] , ... , A[n] :- W (disjoint sets)

(5) B[1] , ... , B[m] :- W (disjoint sets)

(6) X = U(k=1 to n)_A[k] = U(j=1 to m)_B[j]

(7) a[1], ... , a[n] , b[1] , ... , b[m] are positive real numbers

(2) y : W -> [0,oo) is additive

(3) A[1] , ... , A[n] :- W (disjoint sets)

(5) B[1] , ... , B[m] :- W (disjoint sets)

(6) X = U(k=1 to n)_A[k] = U(j=1 to m)_B[j]

(7) a[1], ... , a[n] , b[1] , ... , b[m] are positive real numbers

+(k=1 to n) a[k]*y(A[k]) = +(j=1 to m) b[j]*y(B[j])

page 147 in 1st measure

page 147 in 1st measure

Let W be a s-algebra in X.

Let y : W -> [0,oo] be additive, y(O) = 0.

Let f : X -> [0,oo) be a simple and W-measurable function.

Define the Lebesgue integral of function f with respect to y.

LebS(X,f,dy) = ???

Let y : W -> [0,oo] be additive, y(O) = 0.

Let f : X -> [0,oo) be a simple and W-measurable function.

Define the Lebesgue integral of function f with respect to y.

LebS(X,f,dy) = ???

Since f is simple and W-measurable, there exist:

(1) disjoint sets A[1], A[2], ..., A[n] :- W, with X = U(k=1 to n) A[k]

(2) a[1], a[2], ..., a[n] :- [0,oo)

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

Define LebS(X,f,dy) := +(k=1 to n) a[k]*y(A[k]).

(1) disjoint sets A[1], A[2], ..., A[n] :- W, with X = U(k=1 to n) A[k]

(2) a[1], a[2], ..., a[n] :- [0,oo)

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

Define LebS(X,f,dy) := +(k=1 to n) a[k]*y(A[k]).

Let W be a s-algebra in X.

Let y : W -> |R* be additive.

Let A :- W.

LebS(X,1(A),dy) = ???

Let y : W -> |R* be additive.

Let A :- W.

LebS(X,1(A),dy) = ???

LebS(X,1(A),dy) = y(A)

page 149 in 1st measure

page 149 in 1st measure

Let W be a s-algebra in X.

Let f,g : X -> |R be simple and W-measurable.

Prove that there exist:

(1) E[1], ... , E[n] :- W (disjoint, having union X)

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

such that

(3) f = +(k=1 to n) a[k]*1(E[k])

Let f,g : X -> |R be simple and W-measurable.

Prove that there exist:

(1) E[1], ... , E[n] :- W (disjoint, having union X)

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

such that

(3) f = +(k=1 to n) a[k]*1(E[k])

page 149 in 1st measure

Notice the technical difficulty here.

One day I want to go deeper into this theorem.

MRW 2001.02.24; 2001.06.02

Notice the technical difficulty here.

One day I want to go deeper into this theorem.

MRW 2001.02.24; 2001.06.02