# Properties of probability measures

1. Sep 3, 2004

### 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

2. Sep 3, 2004

### Hurkyl

Staff Emeritus
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.

3. Sep 3, 2004

### Mathechyst

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

Doug

4. Sep 3, 2004

### Hurkyl

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

5. Sep 3, 2004

### Mathechyst

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

Doug

6. Sep 3, 2004

### Hurkyl

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

7. Sep 3, 2004