Consider the unit interval $I=[0,1]$ and let $\mathcal{M}$ be the $\sigma$-algebra of all Lebesgue measurable subsets of $I$. Denote by $m_*$ the Lebesgue outer measure on $\mathcal{M}$. Suppose that $E \subset I$ such that $m_*(E)=1$. Prove that if $A,B \in \mathcal{M}$ and $A\cap E=B\cap E$, then $m(A)=m(B)$.
So I started of using the following idea: $E\backslash A=E\backslash B$. Using the incision property $m(E)-m(A)=m(E)-m(B)$, so $m(A)=m(B)$.
I am not too sure if that is right. I need some clarification on what to do? Is there any other way I could approach this problem?
It suffices to show that $m^\ast(A\cap E) = m(A)$ for all measurable sets $A$, because this will yield $m(A) = m^\ast (A\cap E) = m^\ast (B\cap E) = m(B)$.
To see this, note that $m^\ast (A\cap E) \leq m^\ast (A) = m(A)$, so that we can assume (toward a contradiction) that $m^\ast (A \cap E) < m(A)$.
By definition of $m^\ast$, there is some countable family $(I_n)_n$ of measurable sets (usually intervals, depending on the exact defnition of $m^\ast$) with $A \cap E \subset \bigcup_n I_n$ and such that $\sum_n m(I_n) < m(A)$.
Let $\varepsilon := \frac{1}{2} \cdot (m(A) - \sum_n m(I_n)) > 0$. Then, there is also a countable family $(J_n)$ covering $A^c = [0,1] \setminus A$ such that $\sum_n m(J_n) < m(A^c) + \varepsilon = 1 - m(A) + \varepsilon$.
This shows $$E = (A \cap E) \cup (A^c \cap E) \subset \bigcup_n I_n \cup A^c \subset \bigcup_n I_n \cup \bigcup_n J_n$$
and hence
$$ m^\ast (E) \leq \sum_n m(I_n) + \sum_n m(J_n) < \sum_n m(I_n) + (1 - m(A) + \varepsilon) = 1 -2 \varepsilon + \varepsilon < 1, $$ a contradiction.