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Let A be a collection of sets.

S(R(A)) = ???

S(R(A)) = ???

R(A) c S(A), hence S(R(A) c S(A)

A c R(A) c S(R(A)), hence S(A) c S(R(A))

S(R(A)) = S(A)

see page 35 in 1st measure

A c R(A) c S(R(A)), hence S(A) c S(R(A))

S(R(A)) = S(A)

see page 35 in 1st measure

Let W c P(X). Let u : W -> |R* be additive.

(1) A, B, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(B)?

(1) A, B, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(B)?

Answer: u(B) :- |R

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

Let W c P(X). Let u : W -> |R* be additive.

(1) B, A, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(A\B)?

(1) B, A, A\B :- W

(2) B c A

(3) u(A) :- |R

What can we conclude about the value of u(A\B)?

Answer: u(A\B):-|R and u(A\B) = u(A) - u(B)

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

4) u(A) - u(B) = u(A\B)

Proof:

1) B c A, hence A = B u (A\B) and B n (A\B) = O

2) u is additive, hence u(A) = u(B) + u(A\B)

3) Since u(A) :- |R, both u(B) and u(A\B) are real numbers.

4) u(A) - u(B) = u(A\B)

Let A[n] be a decreasing sequence of sets.

Prove that for every n:-|N

A[n] = A[1] \ ( U(k=1 to k=n-1) A[k]\A[k+1] )

Prove that for every n:-|N

A[n] = A[1] \ ( U(k=1 to k=n-1) A[k]\A[k+1] )

2) page 53 in 2nd measure notebook

1) page 76 in 1st measure notebook

2) is better than 1)

1) page 76 in 1st measure notebook

2) is better than 1)

x does not belong to A\B <=> [ ??? => ??? ]

x does not belong to A\B <=> [ x:-A => x:-B ]

inspiring line:{x does not belong to A\B <=> x:-AcB}

inspiring line:{x does not belong to A\B <=> x:-AcB}

Let A[n] be a sequence of sets.

Prove that

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

Prove that

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

page 78 in 1st measure

Let W c P(X) be a s-ring.

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

Let /\n:-|N A[n] :- W.

Suppose that the series y(A[n]) converges.

What can we conclude about the value of y(lim_sup A[n]) ?

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

Let /\n:-|N A[n] :- W.

Suppose that the series y(A[n]) converges.

What can we conclude about the value of y(lim_sup A[n]) ?

y(lim_sup A[n]) = 0

page 54 in 2nd measure

page 54 in 2nd measure

Let a[n] be a sequence of complex numbers.

Suppose that the series a[n] converges.

What can we immediately conclude?

Suppose that the series a[n] converges.

What can we immediately conclude?

1) a[n] tends to zero

2) lim(k->oo) +(n=k to n=oo) a[n] = 0

2) lim(k->oo) +(n=k to n=oo) a[n] = 0

Let W c P(X) be a s-ring.

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the value y( U(n:-|N) A[n] ) ?

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the value y( U(n:-|N) A[n] ) ?

y( U(n:-|N) A[n] ) = lim(n->oo) y(A[n])

(this limit is proven to exist in the course of the proof)

page 157 in 1st measure

(this limit is proven to exist in the course of the proof)

page 157 in 1st measure

Let W c P(X) be a s-ring.

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the limit of y(A[n]) ?

Let y : W -> |R* be countably additive and y(O)=0.

Let A[n] be an increasing sequence of sets from W.

What can we conclude about the limit of y(A[n]) ?

It exists and

lim(n->oo) y(A[n]) = y( U(n:-|N) A[n] )

hint: A[n] is increasing, hence A[n] = \\*//(k=1 to k=n) A[k]\A[k-1].

page 157 in 1st measure

lim(n->oo) y(A[n]) = y( U(n:-|N) A[n] )

hint: A[n] is increasing, hence A[n] = \\*//(k=1 to k=n) A[k]\A[k-1].

page 157 in 1st measure