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Let A,B c X.

(X\A) \ (X\B) = ???

(X\A) \ (X\B) = ???

(X\A) \ (X\B) = B\A

(B\C) \ (A\D) c ???

(B\C) \ (A\D) c (B\A) u (D\C)

Let F be a s-algebra in X.

Let f : X -> |R*.

Suppose that |f| is measurable.

Does f have to be measurable?

Let f : X -> |R*.

Suppose that |f| is measurable.

Does f have to be measurable?

NO.

Let E c X such that E does not belong to F.

Define f : X -> |R* as follows: f = 1 on E and f = -1 on X\E

{x:-X : f(x)>0} = E does not belong to F

Hence f is not measurable.

But |f| is constant hence measurable.

Let E c X such that E does not belong to F.

Define f : X -> |R* as follows: f = 1 on E and f = -1 on X\E

{x:-X : f(x)>0} = E does not belong to F

Hence f is not measurable.

But |f| is constant hence measurable.

Let R be a ring of sets in X.

Describe K(R) - the algebra generated by R.

Describe K(R) - the algebra generated by R.

K(R) = {EcX : E:-R or X\E:-R}

page 67 in 2nd measure

page 67 in 2nd measure

Let A[1],A[2],A[3],... be a decreasing sequence of sets, having
empty intersection.

What can we conclude from this?

What can we conclude from this?

A[1] = \\*//(n=1 to oo) A[n] \ A[n+1]

/\m:-|N A[m] = \\*//(n=m to oo) A[n] \ A[n+1]

page 97 in 2nd measure

/\m:-|N A[m] = \\*//(n=m to oo) A[n] \ A[n+1]

page 97 in 2nd measure

Let A[1],A[2],A[3],... be a decreasing sequence of sets, having
empty intersection.

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

YES

page 97 in 2nd measure

page 97 in 2nd measure

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

NO

1) Let A[1] = {-7}.

For n>=2, let A[n] = {n,n+1,n+2,n+3,...}.

2) Let A[1]=O, A[2]={7}, A[n] = O, for n>=3

1) Let A[1] = {-7}.

For n>=2, let A[n] = {n,n+1,n+2,n+3,...}.

2) Let A[1]=O, A[2]={7}, A[n] = O, for n>=3

Let A[1],A[2],A[3],... be a decreasing sequence of sets.

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

NO

Let A[n] = {1} for all n:-|N.

Let A[n] = {1} for all n:-|N.

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]

YES

The right side is easily seen to be contained in A[1].

To prove the inverse inclusion one argues by contradiction.

Suppose that x:-A[1] and /\n:-|N ( x:-A[n] => x:-A[n+1] ).

Then it follows that x :- /\n:-|N A[n].

This is a contradiction because the intersection is assumed to be empty.

The right side is easily seen to be contained in A[1].

To prove the inverse inclusion one argues by contradiction.

Suppose that x:-A[1] and /\n:-|N ( x:-A[n] => x:-A[n+1] ).

Then it follows that x :- /\n:-|N A[n].

This is a contradiction because the intersection is assumed to be empty.

Let A[1],A[2],A[3],... be a sequence of sets, having empty
intersection.

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

Suppose that /\n:-|N A[n] c A[1].

Decide if we can conclude that

A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]

NO.

Although we can conclude that A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1],

we cannot conclude that the union is disjoint.

Let A[1] = A[3] = {1}. Let A[2] = A[4] = O. Let A[n] = O for n>4.

Then (A[1]\A[2]) n (A[3]\A[4]) is not empty.

Although we can conclude that A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1],

we cannot conclude that the union is disjoint.

Let A[1] = A[3] = {1}. Let A[2] = A[4] = O. Let A[n] = O for n>4.

Then (A[1]\A[2]) n (A[3]\A[4]) is not empty.