In Mathematical statistics with applications, problem 2.5 they ask:
Show that $(A\cap B)$ and $(A\cap \bar{B} )$ are mutually exclusive, and therefore that $A$ is the union of two mutually exclusive sets, $(A\cap B)$ and $(A\cap \bar{B} )$.
I can show the first part, i.e., if $(A\cap B)$ and $(A\cap \bar{B} )$ are mutually exclusive then $(A\cap B) \cap (A\cap \bar{B} ) = \emptyset$, and $(A\cap B) \cap (A\cap \bar{B} ) = (A\cap A) \cap (B\cap \bar{B} ) = A \cap \emptyset = \emptyset$.
But I don't understand why the authors emphasise that because this is true, then $A$ is the union of the two sets we have shown to be mutually exclusive. Isn't $A$ the union of these two sets, regardless of whether they are mutually exclusive, i.e., from the distributive property, don't we always have:
$$(A\cap B) \cup (A\cap \bar{B}) = A \cap (B \cup \bar{B}) = A\cap \Omega = A$$ even if the two sets are not mutually exclusive?
Best Answer
What author says is that there exist two mutually exclusive sets, say $U$ and $V$, such that $A=U\cup V$. He/she has proved it giving an example: $U=A\cap B$, $V=A\cap \bar B$.