# Question about probability / mutually exclusive events

Decide whether this statement is true or false: Let $$(\Omega, \mathbb{F}, \mathbb{P})$$ be a probability space, if for two events $$A,B \in \mathbb{F}$$ $$\hspace{1cm}$$ $$\mathbb{P}(A \cup B)= \mathbb{P}(A) + \mathbb{P}(B)$$ holds, then $$A \cap B= \emptyset$$

In the solutions it is stated that this statement is false, however I do not really understand why? Isn't the definition that for two mutually exclusive events $$\mathbb{P}(A \cup B)= \mathbb{P}(A) + \mathbb{P}(B)$$ and $$\mathbb{P} (A \cap B) = \emptyset$$?

• Not quite. There are possible events which have probability $0$. Say you are drawing a number $X$ uniformly on $[0,1]$. Let $A$ be the event "$X≤.5$" and let $B$ be the event "$X≥.5$". Then $P(A)=P(B)=\frac 12$ and $P(A\cup B)=1$, but $A\cap B\neq \emptyset$.
– lulu
Jun 29, 2020 at 12:38
• @lulu: thank you for the example Jun 29, 2020 at 13:25

It is always true that $$P(A\cup B)+P(A \cap B)=P(A)+P(B)$$. Hence $$P(A\cup B)=P(A)+P(B)$$ iff $$P(A\cap B)=0$$. But that does not imply that $$A \cap B$$ is the empty set. It can be any event of probability $$0$$.
For example take $$A=[0, \frac 1 2]$$ and $$B=[\frac 1 2 ,1]$$ on $$[0,1]$$ with Lebesgue mesure.
• Thank you for the explanation and the example, so P(A)=0.5=P(B), and $P (A \cup B)= P(A)+P(B)-P(A \cap B)$ $\rightarrow 1=0.5+0.5-P(A \cap B) \rightarrow P(A \cap B)=0$ with $P(A \cap B)=P(0.5)$ but why exactly is $P(0.5)=0$? Jun 29, 2020 at 13:09