Properties of probability measures

[Mathechyst]

I hate it when a fact is so obvious that it isn't obvious how to prove it. Like showing that a subset of a finite set is finite. So ... here goes:

A probability measure P on a \sigma-field \mathcal{F} of subsets of a set \Omega is a function from \mathcal{F} to the unit interval [0,1] such that P(\Omega)=1 and

<br /> P\left(\bigcup_{m=1}^{\infty}A_m\right)=\sum_{m=1}^{\infty}P\left(A_m\right)<br />

for each pairwise disjoint sequence (A_m:m=1,2,3,\ldots) of members of \mathcal{F}. Because P satisfies this summation condition it is said to be countably additive.

The problem is to show that P is finitely additive, that is:

<br /> P\left(\bigcup_{m=1}^{n}A_m\right)=\sum_{m=1}^{n}P\left(A_m\right)<br />

for each pairwise disjoint finite sequence (A_1,\ldots,A_n) of members of \mathcal{F}.

Anyone have any hints to toss my way? Thanks.

Doug

 

[Hurkyl]

Well, you might look at the LHS, and try to consider sets that act as an identity under union, so you can set an infinite union equal to a finite union.

Or, you might look at the RHS and consider numbers that act as an identity under addition, so that you can set an infinite sum equal to a finite sum.

 

[Mathechyst]

The proof would be trivial if P(\emptyset)=0 but that too is a fact that must be proved.

Doug

 

[Hurkyl]

Hrm, do you know any disjoint sequences of sets whose union is the empty set?

 

[Mathechyst]

I would say there's only one: (\emptyset,\emptyset,\ldots).

Doug

 

[Hurkyl]

So what happens if P(\emptyset) \neq 0?

 

[Mathechyst]

Aha. A contradiction.