- #1
Mathechyst
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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 [itex]P[/itex] on a [itex]\sigma[/itex]-field [itex]\mathcal{F}[/itex] of subsets of a set [itex]\Omega[/itex] is a function from [itex]\mathcal{F}[/itex] to the unit interval [itex][0,1][/itex] such that [itex]P(\Omega)=1[/itex] and
[tex]
P\left(\bigcup_{m=1}^{\infty}A_m\right)=\sum_{m=1}^{\infty}P\left(A_m\right)
[/tex]
for each pairwise disjoint sequence [itex](A_m:m=1,2,3,\ldots)[/itex] of members of [itex]\mathcal{F}[/itex]. Because [itex]P[/itex] satisfies this summation condition it is said to be countably additive.
The problem is to show that [itex]P[/itex] is finitely additive, that is:
[tex]
P\left(\bigcup_{m=1}^{n}A_m\right)=\sum_{m=1}^{n}P\left(A_m\right)
[/tex]
for each pairwise disjoint finite sequence [itex](A_1,\ldots,A_n)[/itex] of members of [itex]\mathcal{F}[/itex].
Anyone have any hints to toss my way? Thanks.
Doug
A probability measure [itex]P[/itex] on a [itex]\sigma[/itex]-field [itex]\mathcal{F}[/itex] of subsets of a set [itex]\Omega[/itex] is a function from [itex]\mathcal{F}[/itex] to the unit interval [itex][0,1][/itex] such that [itex]P(\Omega)=1[/itex] and
[tex]
P\left(\bigcup_{m=1}^{\infty}A_m\right)=\sum_{m=1}^{\infty}P\left(A_m\right)
[/tex]
for each pairwise disjoint sequence [itex](A_m:m=1,2,3,\ldots)[/itex] of members of [itex]\mathcal{F}[/itex]. Because [itex]P[/itex] satisfies this summation condition it is said to be countably additive.
The problem is to show that [itex]P[/itex] is finitely additive, that is:
[tex]
P\left(\bigcup_{m=1}^{n}A_m\right)=\sum_{m=1}^{n}P\left(A_m\right)
[/tex]
for each pairwise disjoint finite sequence [itex](A_1,\ldots,A_n)[/itex] of members of [itex]\mathcal{F}[/itex].
Anyone have any hints to toss my way? Thanks.
Doug