1
$\begingroup$

Let ( $ \Omega, \cal A$, P ) be a probability space. Let $(A_n)_n $ be an increasing sequence of events. I am reading a proof using $\sigma$-additivity to prove that $$ P( \cup_n A_n) = \text{lim}_n P(A_n)$$

The proof decomposes $\cup_n A_n$ into non-overlapping sets so that we can apply $\sigma$-additivity.

But actually, I am trying to understand why a proof is needed. The property is "true" for all values of $n$: $P( A_1 \cup A_2 \cup \ A_3 ) = P( A_3)$. Why can't we simply say: "it's true for all values of $n$, therefore it's also true at infinity and therefore $ P( \cup_n A_n) = \text{lim}_n P(A_n)$ is true."?

=== EDIT ===

After reading the comments, I will add this to further pinpoint what I don't understand. Why is the property "$P(A \cup B) = P(A) + P(B)$ for $ A \cap B = \emptyset $" (i.e., no $\sigma$-additivity) insufficient for determining the value of $ P( \cup_n A_n) $?

$\endgroup$
  • 2
    $\begingroup$ If having a property for finite cases implied the same property "at infinity", we wouldn't need sigma-additivity at all... We'd just need finite additivity and we'd say "it works for finite unions so by going at infinity it works for infinite unions". $\endgroup$ – xavierm02 Jan 25 '14 at 9:08
  • $\begingroup$ $1+2+3+...+n$ is finite for each $n$, so $1+2+3+....$ is finite. Does that make sense to you? $\endgroup$ – drhab Jan 25 '14 at 9:30
  • $\begingroup$ What is "true" for all values of $n$? Think about it. And for the record, the theorem which says "we can use induction in mathematics" only applies for "finite $n$", as drhab's comment clearly makes explicit. When you let $n \to \infty$, you usually need extra arguments ; not everything follows. $\endgroup$ – Patrick Da Silva Jan 25 '14 at 9:37
3
$\begingroup$

Because it is false without $\sigma$-additivity.

We can define a probability measure on the set of integers by density. The density of a set of integers is $$ d(A) = \lim_{k \rightarrow \infty} \frac{\#\{A\cap [-k,k]\}}{\#\{[-k,k]\}} $$ where $\#$ is "number of elements" and $[-k,k]$ is the interval of integers. Note that all finite sets have density (probability) zero! density is a finitely additive probability. Now, define $A_k = [-k,k] (A_0=\{0\})$. We have $$ A_k = \{0\} \cup A_1 \cup A_2 \cup \dots \cup A_k $$ is an increasing sequence, $$ \mathbb{N} = \cup_{k=0}^\infty A_k $$ and you can see that your result is not valid. So, it is not valid necessarily without $\sigma$-additivity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.