# probability of the empty set for arbitrary probability measures

I have a probability space $(\Omega, \mathcal{P}(\Omega), P)$. I want to know the probability of the empty set $\{\}$.

Intuitively, I would say this probability is zero. It certainly is for the uniform distribution because there $P(A) = \frac{|A|}{|\Omega|}$.

How do I see this for arbitrary probability measures? I tried $\Omega = \Omega \cup \{\}$ but by $1 = P(\Omega) = P(\Omega \cup \{\}) = P(\Omega) \cdot P(\{\})$ I am led to $P(\{\}) = 1$ which seems very odd.

• $P(\Omega\cup\{\})=P(\Omega)+P(\{\})$. Addition in stead of multiplication. You are dealing with the union of two disjoint sets. Not the intersection of two independent sets. – drhab Dec 18 '14 at 16:32
• In the above (last line), when taking a disjoint union, you get a sum, not a product (product is for intersection of independent events). – Clement C. Dec 18 '14 at 16:32
• Thanks, how could I overlook this ;-) – Marc Dec 18 '14 at 16:36

Hint: Note that $A\cup \emptyset = A$, thus:
$P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A)$, now $P(A)>0 \implies ?$
Hint: $$P(\emptyset)=P(\emptyset\cup\emptyset)=P(\emptyset)+P(\emptyset)$$ This because $\emptyset\cap\emptyset=\emptyset$, i.e. the sets are disjoint.
• What about the other direction: If $P(A)=0$, can I infer $A = \{\}$? That's more a question about conventions because if $P(\{w\})=0$ for a $\omega \in \Omega$, it isn't "needed". But I guess such elements are allowed, aren't they? – Marc Dec 18 '14 at 16:41
• @Marc no you cannot. There are events with probability $0$ that are not empty. For example $P(X=2),X\sim Normal$ – user76844 Dec 18 '14 at 17:11
For any measure space $(X,\mathcal M,\mu)$, $\mu(\varnothing)=0$ by definition. Since a probability measure is just a special case where $\mu(X)=1$, we still have $\mu(\varnothing)=0$.