# Why isn't $P( \cup_n A_n) = \text{lim}_n P(A_n)$ obvious?

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 $P( \cup_n A_n)$?

• 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". – xavierm02 Jan 25 '14 at 9:08
• $1+2+3+...+n$ is finite for each $n$, so $1+2+3+....$ is finite. Does that make sense to you? – drhab Jan 25 '14 at 9:30
• 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. – Patrick Da Silva Jan 25 '14 at 9:37

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.