TITLE: Why is the probability of $\emptyset$ equal to 0?
QUESTION [13 upvotes]: I'm wondering why is the probability of the empty set, $\emptyset$, equal to 0? 
\begin{equation}
P(\emptyset) = 0.
\end{equation}
Isn't the empty set always included when you take the all the possible subsets of a set?

REPLY [5 votes]: I think a more rigorous proof than previously given as answers uses more axioms of a probability measure. Of the three axioms, we need:

$\mathrm{P}(\Omega)=1$.

Given $A_i$ are mutually disjoint events in a sample space $\Omega$, $\mathrm{P}\left(\cup_{i=1}^{\infty}A_i\right)= \sum_{i=1}^{\infty}\mathrm{P}(A_i)$


So we say $A_i = \emptyset$, because these are mutually disjoint, we can use axiom 2, such that:
$$\mathrm{P}\left(\cup_{i=1}^\infty A_i\right)=\sum_{i=1}^{\infty}\mathrm{P}(A_i)=\mathrm{P}(\emptyset)+P(\emptyset)+\dotsb.$$
However, we know from axiom one that the probability of a event must be finite from axiom 1. Therefore $\mathrm{P}(\cup_{i=1}^\infty A_i)<\infty$ is true only if $\mathrm{P}\left(\emptyset\right)=0$.