Finite additivity in outer measure Let $\{E_i\}_{i=1}^n$ be finitely many disjoint sets of real numbers (not necessarily Lebesgue measurable) and $E$ be the union of all these sets. Is it always true that 
$$
m^\star (E)=\sum_{i=1}^N m^\star(E_n)
$$
where $m^\star$ denotes the Lebesgue outer measure? If not, please give a counterexample. The Vitali set is a counterexample in the countable case, but I am not sure whether it is false in finite case.

The sentence "$m^{*}$ is not finitely additive" is independent from the theory $ZF+DC$.
Fact 1. (ZF)+(AC). $m^{*}$ is not finitely additive.
Proof
Let $X$ be a subset of the real line $R$. We say that $X$ is a Bernstein set in $R$ if
for every non–empty perfect set $P \subseteq R$ both sets
$P \cap X$, $P \cap (R \setminus X)$
are non–empty. 
It is well known that there exists a Bernstein subset $X$ of $R$.
Let consider two sets $[0,1]\cap X$ and $[0,1]\setminus X$. Then 
(i) $m_{*}([0,1]\cap X)=m_{*}([0,1]\setminus X)=0$;
(ii) $m^{*}([0,1]\cap X)=m^{*}([0,1]\setminus X)=1$.
Indeed, since both $[0,1]\cap X$ and $[0,1]\setminus X$ (like $X$ and $R \setminus X$) no contain any non–empty perfect subset
we claim that that (i) holds true.  
If we assume that $m^{*}([0,1]\cap X)<1$ then there will be a non–empty perfect subset $Y \subset [0,1]$ such that $Y \cap ([0,1]\cap X)=\emptyset$. The later relation  implies  that $R \setminus X$ like $[0,1]\setminus X$ contains a non–empty perfect set and we get a contradiction that $X$ is a Bernstein subset of $R$.
The proof of the equality $m^{*}([0,1]\setminus X)=1$ is similar.
Hence we have 
$$m^{*}(([0,1]\cap X))\cup ([0,1]\setminus X))=m^{*}([0,1])=1 <1+1=m^{*}([0,1]\cap X)+m^{*}([0,1]\setminus X).
$$
Fact 2. (Solovay model= (ZF)+(DC)+(every subset of $R$ is Lebesgue measurable)). $m^{*}$ is finitely additive.
Proof. In Solovay model $m^{*}=m$. Hence  $m^{*}$ is finitely additive since $m$ is finitely additive.
