1. ## lebesque measurablity

$m{\ast}A=0{\Rightarrow}m{\ast}(A{\cup}B)=m{\ast}B$
how can ı show this?

2. Hello,
Originally Posted by sah_mat
$m{\ast}A=0{\Rightarrow}m{\ast}(A{\cup}B)=m{\ast}B$
how can ı show this?
$m{\ast}(A \cup B)+m {\ast}(A \cap B)=m{\ast}(A)+m{\ast}(B)$

but $m{\ast}(A \cap B) \leq m{\ast}(A)=0$ because $A \cap B \subseteq A$

Since a measure is always positive, $m{\ast}(A \cap B)=0$

Hence $m{\ast}(A \cup B)=m{\ast}(A)+m{\ast}(B)=m{\ast}(B)$

3. thanks moo, ı have another question here;
if $E{\displaystyle1} and E{\displaystyle2}$ are measurable than $m(E1{\cup}E2) +m(E1{\cap}E2)=m(E1)+m(E2)$

ı have an exam tomorrow so please answer as soon as possible