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 $\mathbb{P} (A \cap B) = \emptyset$?
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.