# Countable additivity and $P(\Omega)=1$

The first axiom of probability states that $P(\Omega)=1$. On the other hand, we have that $$P\left(\bigcup_{i = 1}^\infty E_i\right) = \sum_{i=1}^\infty P(E_i)$$ where $E_i$ are pairwise disjoint events.

By the first axiom, it's necessary that finite sum of probabilities of pairwise disjoint events equals $1$. Now, everything is perfectly fine as long as the probabilities add up to $1$. According to 3rd axiom, we can extend this to countably many disjoint events. I think there's a problem though - can you show me a countably infinite sequence of non-negative numbers that, when added together, equal $1$?

Ok, here it is: $\sum_n2^{-n}$, $n\in \mathbb{N} \setminus \{0\}$.

But it's not exactly equal to $1$, it's $1$ in the limit, which is I believe quite different. When we add countably many numbers in the summation above, we will quite likely get arbitrarily close to $1$, but we won't reach $1$.

I know that infinity is not an easy topic, but could you clarify this for me?

• Stunning fact: the sum of the series $\sum\limits_{n=1}^\infty2^{-n}$ is exactly equal to $1$. – Did Oct 20 '15 at 10:36
• This seems similar to the common misconception that $0.999... \neq 1$. It is true, that for any partial sum $\sum_{n=1}^{k}2^{-n}$ we get a number less than $1$, but in the limit we do in fact get exactly $1$. And a common reply to the argument that it's 'something' less than $1$ would be to ask what that number is. – Santeri Oct 20 '15 at 10:41

The notation $$\sum_{i=1}^\infty \cdots$$ always means the limit of the partial sums; there's nothing else for it to mean. So it's a misunderstanding to claim that $\sum_{i=1}^\infty\cdots$ lacks some property that the limit would have because taking the limit is implicit in that notation.
Thus, $\sum_{i=1}^\infty 2^{-i}$ is exactly equal to $1$, because $1$ is the limit and "the limit" is what $\sum_{i=1}^\infty 2^{-i}$ means.