Prove that if $E$ is measurable, then for any sets $A$, $m^*(E \cap A) + m^*(E^c \cap A) = m^*(A)$ 
Prove that if $E$ is measurable,then for any sets $A$, $m^*(E \cap A) + m^*(E^c \cap A) = m^*(A)$ 

I managed to prove the converse. (this question was edited)
This is not a duplicate from the following post because we use different definition of measurabiliy/outer measure and also because my setting is nore general, i.e. I am considering in the general $\mathbb R^d$ rather than just the real line.
Equivalent Definition of Measurable set
First the definitions:
If $E$ is any subset of $\mathbb R^d$, the outer measure of $E$ is $m^*(E) = inf \sum_{j \in \mathbb N} |Q_j|$ where the infimum is taken over all countable subcoverings $E  \subset \cup_{j \in \mathbb N}Q_j$ by CLOSED cubes. 
We say that $E$ is measurable if for any $\epsilon >0$ there is an open set $O$ containg $E$ such tha $m^*(O - E) \leq \epsilon$. If $E$ is measurable, we define the measure of $E$ to be $m(E) =m^*(E)$.
(I am using the textbook Real Analysis: book3 by Stein and Shakarchi from Princeton)
We know that by countable subadditivity of outer measure $m^*(E \cap A) + m^*(E^c \cap A) \geq m^*(A)$. But I am not sure how to prove the other direction of the inequality.
 A: There are some steps in the proof


*

*Show that the equality $m_*(E \cap A) + m_*(E^c \cap A) = m_*(A)$ 
is valid if $E$ is open, $E=U$. 

*Show the equality for any measurable $E$ ( that is, any $E$ that can be approximated by open sets).
If you have 1. then 2. is not hard and it's a good exercise.
To prove 1., first prove the following: 


*If $A$, $B$ are subsets of $\mathbb{R}^n$ and $d(A,B)>0$ then $m_{\star}(A)+m_{\star}(B)=m_{\star}(A\cup B)$


This is the so called "Caratheodory condition" for the outer measure $m_{\star}$. It's a good exercise to show it for this outer measure.
Let's show that 3. implies 1. This is the only tricky part. The ideas come from Caratheodory I suppose.
Take $U$ open. If $U$ is the whole space then 1. holds for any $A$. So assume that $U$ is not the whole space. Note that $U=\{x\ \ | \ d(x, U^{c})>0\}$. Consider the following partition of $U$ into layers $L_n$ where
$$L_n= \{x\ | \ \frac{1}{n-1}>d(x,U^{c})\ge \frac{1}{n} \}$$ 
for all $n\ge 1$. It's easy to see that $L_n$ is a partition of $U$. Moreover, we have $d(L_m,L_n)>0$ for $|m-n|>1$. 
Let now $A$ be a subset of $\mathbb{R}^n$. We want to show that 
$$m_{\star}(U\cap A) +m_{\star}(U^c\cap A)=m_{\star}(A)$$
If $m_{\star}(U\cap A)= \infty$ this is clear. Assume that $m_{\star}(U\cap A)<\infty$. 
Note that $d( L_{m}\cap A,  L_n\cap A)>0$ for $|m-n|>1$. Therefore we have for every $N$ ( Caratheodory condition and easy induction)
$$\sum_{k=1}^N m_{\star}( L_{2k-1}\cap A) = m_{\star}((\cup_{k=1}^N L_{2k-1})\cap A)\le m_{\star}(U\cap A) < \infty\\
\sum_{k=1}^N m_{\star}(L_{2k}\cap A) = m_{\star}((\cup_{k=1}^N L_{2k})\cap A)\le m_{\star}(U\cap A) < \infty$$
From here we conclude that the series $\sum m_{\star}( L_{2n-1}\cap A)$, $\sum m_{\star}( L_{2n}\cap A)$ are convergent and so $\sum m_{\star}(L_{n}\cap A)$ is convergent
. 
Therefore, if $F_n=\{x \ | \ d(x,U^c )\ge \frac{1}{n}\}$, $V_n= U\backslash F_n$, we have 
$$m_{\star}(  V_n\cap A)\to 0$$ and so 
$$m_{\star}(  F_n\cap A) \to m_{\star}(  U\cap A)$$
Now we have 
$$m_{\star}( A )\ge m_{\star}( ( F_n\cap A)\cup ( U^c\cap A))= m_{\star}(  F_n\cap A)+ m_{\star}(  U^c\cap A)$$ and passing $n$ to $\infty$ we get 
$$m_{\star}( A )\ge m_{\star}( U \cap A)+ m_{\star}(  U^c\cap A)$$
