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

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    $\begingroup$ 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 $A\cap B\neq \emptyset$. $\endgroup$ – lulu Jun 29 at 12:38
  • $\begingroup$ @lulu: thank you for the example $\endgroup$ – xxDianaxx Jun 29 at 13:25
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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.

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  • $\begingroup$ 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 $P(0.5)=0$? $\endgroup$ – xxDianaxx Jun 29 at 13:09
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    $\begingroup$ That is a property of Lebesgue measure. $\endgroup$ – Kavi Rama Murthy Jun 29 at 13:12
  • $\begingroup$ thank yo, will look it up $\endgroup$ – xxDianaxx Jun 29 at 13:25

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