Let $\mathcal{M} \subset \mathcal{P}(\mathbb{R})$ such that: $$ \mathcal{M} = \{A \subset \mathbb{R}, \forall E \subset \mathbb{R}, \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c) \} $$ $\mu^*$ is the Lebesgue measure. Let $\mathcal{A}$ be an algebra on $\mathbb{R}$.
Show that $\mathcal{A} \subset \mathcal{M}$.
I previously had shown that if $\mu$ is a measure on $(\mathbb{R}, \mathcal{A})$ then $\forall A \in \mathcal{A}, \mu(A)= \mu^*(A)$.
Now in order to show that $\mathcal{A} \subset \mathcal{M}$ I have to show that if $A \in \mathcal{A}$ then $\forall E \subset \mathbb{R}$, $\mu^*(E) = \mu^*(A \cap E) + \mu^*(A^c \cap E).$
I tried to to do it by double inequality. It is true that $E \subset E \cap A \cup E \cap A^c$ thus $\mu^*(E) \leq \mu^*(E\cap A) + \mu^*(E \cap A^c)$, but I am unable to show the inequality the other way.