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

Can we conclude that {A[n]\A[n+1]}(n:-|N) is a disjoint collection?

Can we conclude that {A[n]\A[n+1]}(n:-|N) is a disjoint collection?

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

\/k:-|N {A[n]\A[n+1]}_(n=k to n=oo) is disjoint.

Decide if we can conclude that

\/k:-|N {A[n]\A[n+1]}_(n=k to n=oo) is disjoint.

NO

(1) Let A[2n]={1}, A[2n+1]=O.

//Let A[2n] = {2n, 2n+1, 2n+2, 2n+3, ...}. Let A[2n+1] = {2n}.

//Then //\\(n:-|N) A[n] = O. And for every even n:-|N,

//(A[n]\A[n+1]) n (A[n+2]\A[n+3]) = {n+1,n+2,n+3,n+4,...} n {n+3,n+4,...} != O

(1) Let A[2n]={1}, A[2n+1]=O.

//Let A[2n] = {2n, 2n+1, 2n+2, 2n+3, ...}. Let A[2n+1] = {2n}.

//Then //\\(n:-|N) A[n] = O. And for every even n:-|N,

//(A[n]\A[n+1]) n (A[n+2]\A[n+3]) = {n+1,n+2,n+3,n+4,...} n {n+3,n+4,...} != O

Let B[1],B[2],B[3],... be an increasing sequence of sets.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

YES

page 98 in 2nd measure

page 98 in 2nd measure

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

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

YES

The right-hand set is easily seen to be contained in the left- hand set.

Take any x belonging to the left-hand set. We have (1) \/n:-|N x:-B[n].

Let's argue by contradiction. Suppose that (2) /\n:-|N ( x:- B[n] => x:-B[n-1] ).

The right-hand set is easily seen to be contained in the left- hand set.

Take any x belonging to the left-hand set. We have (1) \/n:-|N x:-B[n].

Let's argue by contradiction. Suppose that (2) /\n:-|N ( x:- B[n] => x:-B[n-1] ).

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

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

Decide if we can conclude that

\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],

where B[0] = O.

NO

We can indeed conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1]

but the right-hand union does not have to be disjoint.

We can indeed conclude that

\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1]

but the right-hand union does not have to be disjoint.

Let B[1],B[2],B[3],... be an increasing sequence of sets.

Decide if we can conclude that

/\n:-|N B[n] = \\*//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

Decide if we can conclude that

/\n:-|N B[n] = \\*//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

YES

page 42 in 1st measure

page 42 in 1st measure

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

Decide if we can conclude that

/\n:-|N B[n] = \\//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

Decide if we can conclude that

/\n:-|N B[n] = \\//(k=1 to k=n) B[k]\B[k-1],

where B[0] = O.

NO

Let B[1] = {7}, B[2] = O.

Then B[2] != B[1] u B[2]\B[1].

We can only conclude:

/\n:-|N B[n] c \\//(k=1 to k=n) B[k]\B[k-1],

Let B[1] = {7}, B[2] = O.

Then B[2] != B[1] u B[2]\B[1].

We can only conclude:

/\n:-|N B[n] c \\//(k=1 to k=n) B[k]\B[k-1],

Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.

Suppose that both sequences a[n] and b[n] are increasing.

What can we conclude?

Suppose that both sequences a[n] and b[n] are increasing.

What can we conclude?

There exist a,b:-[0,oo] such that

(1) lim a[n] = a

(2) lim b[n] = b

(3) lim a[n]*b[n] = a*b

page 93 in 2nd measure

(1) lim a[n] = a

(2) lim b[n] = b

(3) lim a[n]*b[n] = a*b

page 93 in 2nd measure

Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.

Let a,b:-[0,oo].

(1) lim a[n] = a,

(2) lim b[n] = b.

Decide if we can conclude that

(3) lim a[n]*b[n] = a*b.

Let a,b:-[0,oo].

(1) lim a[n] = a,

(2) lim b[n] = b.

Decide if we can conclude that

(3) lim a[n]*b[n] = a*b.

NO

1) a[n] = 1/n; b[n] = n

2) a[n] = 1/n; b[n] = n^2

1) a[n] = 1/n; b[n] = n

2) a[n] = 1/n; b[n] = n^2

Let F be a s-algebra in X.

Let f : X -> |R* be measurable.

What can we conclude about the measurability of 1/f ?

Let f : X -> |R* be measurable.

What can we conclude about the measurability of 1/f ?

It is measurable, no matter how we define: 1/0, 1/oo, 1/-oo.

page 93 in 2nd measure

page 93 in 2nd measure