Prove: If $E_1, E_2$ are measurable then $E_1\cup E_2$ are measurable
It sufficient to show that for all $A$ :
$$m^* (A)\geq m^*(A\cap (E_1\cup E_2))+m^*(A\cap (E_1\cup E_2)^C)$$
Now $A\cap (E_1\cup E_2)^C=A\cap (E_1^C\cap E_2^C)=(A\cap E_1^C)\cap( A\cap E_2^C)=A\cap E_1^C\cap E_2^C$
And $A\cap (E_1\cup E_2)=(A\cap E_1) \cup (A \cap E_2)$
How can I prove the inequality?
Best Answer
We have
\begin{align} \quad m^*(A) &= m^*(A \cap E_1) + m^*(A \cap E_1^c) \\ &= m^*(A \cap E_1) + m^*([A \cap E_1^c] \cap E_2) + m^*([A \cap E_1^c] \cap E_2^c) \\ &= m^*(A \cap E_1) + m^*(A \cap E_1^c \cap E_2) + m^*(A \cap [E_1 \cup E_2]^c) .\\ & \end{align}
To conclude, note that $$A\cap(E_1\cup E_2)= (A \cap E_1) \cup (A \cap E_1^c \cap E_2)$$
and that $m^*$ is subadditive.