Proving Properties of Countable Sets & Probability Spaces

[hellbike]

1.prove that for any set X: |X|<c <=> in P(X) exist such countable set family F, that sigma algebra generated by F contains all points.

2.let (X,E,u) be probability space and A_1,...,A_2009 in E have property u(A_i)>=1/2. Prove that there exist x such is in A_i for atleast 1005 different i.

i need some tip in solving this.

 

[Citan Uzuki]

On problem 1, for the forward implication, note that if |X| ≤ c, then you can assume WLOG that X⊆R. For the reverse, show that the cardinality of the Σ-algebra generated by any countable family of sets is at most the continuum.

For problem 2, try applying a counting method. Specifically, define S_n as the set of all points that are in exactly n of the A_i. What is ∑nμ(S_n)?

 

[hellbike]

What do you mean by counting method? Induction?

And this sum is between 1/2 and 1, but i don't know why this would be useful?

and for problem 1:
->
is this sufficient to say that set of (p,q) for p,q in Q can do the job for any subset of R?

 

[Citan Uzuki]

What do you mean by counting method? Induction?

I mean extending combinatorial methods to the measure-theoretic setting. In this case, we want to "count" (i.e. find the measure of) the set of points in the A_i with multiplicity (i.e. a point that appears in two different A_i would be counted twice) in two different ways. First, we divide it up into the disjoint sets S_n, and observe that each point in S_n will be counted exactly n times, so the total measure (with multiplicity) is \sum_{n=1}^{2009}n\mu (S_n). On the other hand, each point is counted once each time it appears in an A_i, so the total measure (with multiplicity) is \sum_{i=1}^{2009}\mu (A_i). So these two sums are equal. Can you see how this implies that μ(S_n)>0 for at least one n≥1005?

is this sufficient to say that set of (p,q) for p,q in Q can do the job for any subset of R?

Nearly, although since you're generating a Σ-algebra on X, you should write (p,\ q) \cap X,\ p,\ q\in\mathbb{Q}

 

[hellbike]

yes, thank you very much.