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

$$P\left(\bigcup_{m=1}^{\infty}A_m\right)=\sum_{m=1}^{\infty}P\left(A_m\right)$$

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:

$$P\left(\bigcup_{m=1}^{n}A_m\right)=\sum_{m=1}^{n}P\left(A_m\right)$$

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.