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

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

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

Suppose that y(A[k]) :- |R, for some k:-|N.

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

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

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

Suppose that y(A[k]) :- |R, for some k:-|N.

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

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

the limit is shown to exist in the course of the proof

In the proof, we use the previous item.

page 158 in 1st measure

page 55 in the 2nd measure notebook

the limit is shown to exist in the course of the proof

In the proof, we use the previous item.

page 158 in 1st measure

page 55 in the 2nd measure notebook

Let W c P(X) be a ring.

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = oo.

What can we conclude?

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = oo.

What can we conclude?

y(B) = oo

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(A) = oo, so is y(B).

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(A) = oo, so is y(B).

Let W c P(X) be a ring.

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = -oo.

What can we conclude?

Let y : W -> [-oo,oo] be additive.

Let A,B :- W.

Suppose that A c B and y(A) = -oo.

What can we conclude?

y(B) = -oo

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + Y(B\A).

Since y(A) = -oo, so is y(B).

A c B, hence B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + Y(B\A).

Since y(A) = -oo, so is y(B).

Let W c P(X) be a ring.

Let y : W -> [0,oo] be additive.

Let A,B :- W.

Suppose that A c B.

What can we conclude?

Let y : W -> [0,oo] be additive.

Let A,B :- W.

Suppose that A c B.

What can we conclude?

y(A) <= y(B)

Since A c B, B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(B\A)>=0, we get y(A) <= y(A) + y(B\A) = y(B).

Hence y(A) <= y(B).

Since A c B, B = A u (B\A) disjointly.

Since y is additive, y(B) = y(A) + y(B\A).

Since y(B\A)>=0, we get y(A) <= y(A) + y(B\A) = y(B).

Hence y(A) <= y(B).

Let W be a ring of sets.

Let y : W -> [0,oo] be countably additive.

Can we conclude that y is countably subadditive on W?

Let y : W -> [0,oo] be countably additive.

Can we conclude that y is countably subadditive on W?

Yes.

hint: U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ U(k=1 to n-1)_A[k] ]

page 84 in 1st measure

hint: U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ U(k=1 to n-1)_A[k] ]

page 84 in 1st measure

Let W be a ring of sets.

Let y : W -> [0,oo[ be additive.

Prove that

/\B,A:-W |y(A) - y(B)| <= y(A+B)

Let y : W -> [0,oo[ be additive.

Prove that

/\B,A:-W |y(A) - y(B)| <= y(A+B)

We will use [ E c F => y(E) <= y(F). ]

y(A) <= y(A u B) = y( (A+B) u (AnB) ) = y(A+B) + y(AnB) <= y(A+B) + y(B)

Hence y(A) <= y(A+B) + y(B).

Since the values are real numbers, we get y(A) - y(B) <= y(A+B).

y(A) <= y(A u B) = y( (A+B) u (AnB) ) = y(A+B) + y(AnB) <= y(A+B) + y(B)

Hence y(A) <= y(A+B) + y(B).

Since the values are real numbers, we get y(A) - y(B) <= y(A+B).

Let W be a semi-ring of sets. Let y : W -> [-oo,oo] be finitely
additive.

Let E be written, in two different ways, as a finite disjoint union of sets in W.

What can we conclude?

Let E be written, in two different ways, as a finite disjoint union of sets in W.

What can we conclude?

Let E = U(i=1 to n) C[i] = U(j=1 to m) D[j]. And C[i],D[j] :- W.

We can conclude:

+(i=1 to n) y(C[i]) = +(j=1 to m) y(D[j])

page 86 in 1st measure

Notice here that finite additivity is required.

We can conclude:

+(i=1 to n) y(C[i]) = +(j=1 to m) y(D[j])

page 86 in 1st measure

Notice here that finite additivity is required.

Let J be a semi-ring of sets.

Let y : J -> [-oo,oo] be finitely additive.

Extend function y so that y : R(J) -> [-oo,oo] is additive.

Let y : J -> [-oo,oo] be finitely additive.

Extend function y so that y : R(J) -> [-oo,oo] is additive.

page 87, 88 in 1st measure

Notice that finite additivity is required.

Notice that finite additivity is required.

Let a,b,c belong to [0,oo].

Prove that (a+b)*c = (a*c) + (b*c).

Prove that (a+b)*c = (a*c) + (b*c).

___

Let C and D be sets.

Suppose that

1) (a,b) :- C x D

2) (x,y) :- C x D

What can we conclude about (a,y) and (x,b) ?

Suppose that

1) (a,b) :- C x D

2) (x,y) :- C x D

What can we conclude about (a,y) and (x,b) ?

(a,y) :- C x D

(x,b) :- C x D

(x,b) :- C x D