Let $A$ and $B$ are sets measurables. Prove $\mu^*(A \cup B) + \mu^*(A \cap B) \leq \mu^*(A) + \mu^*(B)$
I tried to use $A \cap B \subset A \subset A \cup B$. Similarly to $B$ and I apply the Lebesgue measure but that doesn't work. Could you help me?
HINT:
Use the following properties of the Lebesgue measure and exterior measure
$$\mu^{\star}(A) = \inf_{U \text{meas}, A\subset U} \mu(U)$$
Let $\epsilon >0$. Take $U$, $V$ measurable so that $A\subset U$, $B\subset V$ and $$\mu(U)\le \mu^{\star}(A)+ \epsilon/2$$ $$\mu(V)\le \mu^{\star}(B)+ \epsilon/2$$ Now $ A\cup B\subset U\cup V$ and $ A\cap B\subset U\cap V $. So $\mu^{\star}(A\cup B) \le \mu(U\cup V)$, $\mu^{\star}(A\cap B) \le \mu(U\cap V)$. Therefore $$\mu^{\star}(A\cup B) + \mu^{\star}(A\cap B) \le \mu(U\cup V) + \mu(U\cap V) = \mu(U) + \mu(V) \le \mu^{\star}(A) + \mu^{\star}(B)+\epsilon$$