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How to prove that finite additivity follows from countable additivity!!?

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Could you be more specific on additivity of what? – user17762 Nov 13 '10 at 17:42
Additivity with probabilities – Osama Gamal Nov 13 '10 at 17:53
up vote 10 down vote accepted

I guess this question arises because in the definition of countable additivity it is assumed that there are infinitely many events. But if there are finitely many events, you can just throw in countably many empty events. That is, given $A_1,\ldots,A_n$, define $A_{n+1}=\emptyset$, $A_{n+2}=\emptyset$, and so on, then apply countable additivity.

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I assume that you want to show that finite additivity follows from countably infinite additivity. In general, though countable means both finite and countably infinite.

We are given that $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.

So we now want to show $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$, $\forall n \in \mathbb{N}$, whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.

Now consider $A_{i} = \emptyset$, $\forall i > n$.

By the way we have defined, note that $A_i \cap A_j = \emptyset$, $\forall i \neq j$. (Note: $A \cap \emptyset = \emptyset$, $\forall A$).

By countably infinite additivity, $\mu(\displaystyle \cup_{i=1}^\infty A_i) = \displaystyle \sum_{i=1}^\infty \mu(A_i)$ and note that $\mu(A_i) = 0$, $\forall i > n$.

Also note that $\displaystyle \cup_{i=1}^\infty A_i = \displaystyle \cup_{i=1}^n A_i$, since $A_i = \emptyset$, $\forall i > n$.

Hence, $\mu(\displaystyle \cup_{i=1}^n A_i) = \displaystyle \sum_{i=1}^n \mu(A_i)$ whenever $A_i \cap A_j = \emptyset$, $\forall i \neq j$.

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The empty set is not usually denoted by $\phi$. Instead, I recommend $\emptyset$ or ∅. – Jonas Meyer Nov 13 '10 at 18:32
Whether "countable" means "countably infinite" or "at most countably infinite" depends on who you ask. E.g. says the former. I do prefer the meaning you use, as it is the "common sense" approach. – Jonas Meyer Nov 13 '10 at 18:40
@ Jonas: Accepted. Changed $\phi$ to $\emptyset$ – user17762 Nov 13 '10 at 19:16
@ Jonas: I am unable to access the link you provided in your comment. – user17762 Nov 13 '10 at 19:18
@Sivaram: I don't know how to make the link work for you, but it is to the Springer online encyclopedia entry for "Countable set". The body reads simply, "A set of the same cardinality as the set of natural numbers. For example, the set of rational numbers or the set of algebraic numbers." I don't know how common the convention is, but it is also mentioned in Royden's Real Analysis: "Many authors restrict the use of the word 'countable' to sets which are infinite and countable, but our definition includes the finite sets among the countable sets," page 10 of the 2nd edition. – Jonas Meyer Nov 13 '10 at 19:26

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