Let $m_*(A)$ denote the Lebesgue outer measure, and when $A$ is measurable, let $m(A)=m_*(A)$ be the Lebesgue measure.
Let $U=\{E_1,\cdots,E_N\}$ be a finite collection of pairwise disjoint Lebesgue measurable sets (of $\mathbb{R}^n$), and denote $E:=\bigcup_{i=1}^{N} E_i$. We know that (finite additivity): $$ m(E)=\sum_{i=1}^{N}m(E_i)\tag{1}$$ If we add another set $A$, which may not be measurable, is it true that $$m_*(A\cap E)=\sum_{i=1}^{N}m_*(A\cap E_i)\tag{$*$}$$ (1) is a special case of ($*$) when $A=E$. Anyone have any idea on proving ($*$)?
I thought $m_*(A\cap E)=m_*(A)-m_*(A\setminus E)$ may be useful but I don't know how to proceed.
Note that the definition of a set $E$ is measurable is that $m_*(A)=m_*(A\setminus E)+m_*(A\cap E)$ for all $A$.
Consider the case $U = \{E_1, E_2\}$. You can easily get the general statement by induction. Because $E_1$ is measurable, and using $A \cap (E_1 \cup E_2) $ as a "test" set, we have:
\begin{align} m_*(A \cap (E_1 \cup E_2) \cap E_1 ) &= m_*(A \cap (E_1 \cup E_2) ) - m_*(A \cap (E_1 \cup E_2) \setminus E_1 ) \\ &=m_*(A \cap (E_1 \cup E_2) ) - m_*(A \cap E_2 ). \end{align}
On the other hand $m_*(A \cap (E_1 \cup E_2) \cap E_1 ) = m_*(A \cap E_1)$.