# Does it follow that for disjoint sets $P(X \cup Y) = P(X) + P(Y)$ that for any sets $P(X \cup Y) + P(X \cap Y) = P(X) +P(Y)$?

This is a question about probability theory. One often sees a probability measure characterised as being additive "for disjoint $X,Y$ we have: $P(X \cup Y) = P(X) + P(Y)$ " but sometimes as being additive like so: $P(X \cup Y) + P(X \cap Y) = P(X) +P(Y)$ . I guess what I'm asking is, how do you prove the inclusion/exclusion version of additivity from the version for disjoint sets. (The converse is trivial).

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Split $X$ into the disjoint union of $X \cap Y$ and some other set $A$, and split $Y$ into the disjoint union of $X \cap Y$ and some other set $B$.
$X = (X\setminus Y) \cup (X\cap Y)$, therefore $P(X) = P(X\setminus Y) + P(X\cap Y)$.
On the other hand, $X\cup Y = (X\setminus Y)\cup Y$, and $Y\cap (X\setminus Y) = \emptyset$, therefore: $$P(X) + P(Y) = P(X\setminus Y)+P(X\cap Y)+P(Y) = P(X\setminus Y \cup Y) + P(X\cap Y) = P(X\cup Y)+P(X\cap Y)$$