[Math] Prove if $E_1$ and $E_2$ are measurable then $m(E_1\cup E_2)+m(E_2\cap E_2)=m(E_1)+m(E_2)$

measure-theory

by additivity
$m(E_1\cup E_2)=m(E_1)+m(E_2)$ (because $E_1,E_2$ are measurable)
but i don't know what to do with $E_1\cap E_2$.
I tried to use demorgan's identity to solve this part but this is not clear to me.

Best Answer

Firstly, $m(E_1 \cup E_2) = m(E_1) + m(E_2)$ is not true in general. Its only true if $E_1 \cap E_2 = \emptyset$.

So, with that cleared up, note that there are sets $A, B$ and $C$ such that $A\cap B = A\cap C = B\cap C = \emptyset$ and $E_1 = A \cup B,\ E_2 = B\cup C$ and $E_1\cap E_2 = B$.

Therefore $E_1\cup E_2 = A\cup B\cup C$ and using the fact that we have additivity for disjoint sets, we can conclude $$\begin{align*}m(E_1 \cup E_2) + m(E_1\cap E_2) &= m(A\cup B\cup C) + m(B)\\ &= m(A) + m(B) + m(C) + m(B)\\ &= m(A\cup B) + m(B\cup C)\\ &= m(E_1) +m(E_2).\end{align*}$$

So, big important thing to remember, what you have in general is$$ m\left(\bigcup_i A_i\right) \leq \sum_i m(A_i)$$with equality only holding in the case of disjoint sets.

Related Question