Lemma: Let $A\_1,\ldots,A\_n$ are events $n\in\mathbb{N}$ then 
$$
\sum\_{i=1}^n \mathbb{P}(A\_i) = \mathbb{P}(\cup\_{i=1}^n A\_i)
$$
if and only if $A\_1,\ldots,A\_n$ are mutually exclusive.
Both ways are shown by an easy induction.  

However, I think that we are assuming that the probability spaces are finite. Does this lemma still hold if we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\mathcal{F}$ is countable or uncountably infinite.
Thanks in advance.



Equality in the union bound. 
Your "lemma" is false for finite probability spaces, e.g., 
$\Omega = \{a,b\}, \mathbb P(\{a\})=0,\mathbb P(\{a,b\})=1, \mathbb P(\{a\} \cup \{a,b\})=1.$ 


After you fix it, a cannon to swat the fly is [inclusion-exclusion](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle), or more specifically, the [Bonferroni inequalities](http://en.wikipedia.org/wiki/Boole's_inequality).


I think people are confusing your question as stated with the natural and very elementary question of whether countably additive probabilities must be uncountably additive, and the example of Lebesgue measure on $[0,1]$ shows this this is not the case. 


You should very rarely do anything with the sample space itself in any intrinsic probability question. See the answer gowers gave to [this question](https://mathoverflow.net/questions/12516/a-random-variable-is-it-a-function-or-an-equivalence-class-of-functions-closed) and this Tao [blog entry](http://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/). It's ok to have a sample space when you apply probability to something like an analysis question (e.g., proving the Weierstrauss approximation theorem using probability) or use the probabilistic method in combinatorics. 


