Let u be a measure on a sigma algebra A of subsets of X. Prove that u(E)+u(F)=u( )+u( )
$(E \cup F) = E \cup (F \cap \overline{E})$
These are disjoint so
$u(E \cup F) =u(E) + u(F \cap \overline{E})$
$u(E \cup F) + u(E \cap F) = u(E) + u(F \cap \overline{E}) + u(E \cap F)$
$F=(F \cap E) \cup (F \cap \overline{E})$
and both of these are disjoint so
$u(F) = u(F \cap \overline{E}) + u(E \cap F)$
so
$u(E \cup F) + u(E \cap F)=u(E) + u(F)$