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.
Finite Additivity in Outer Measure – Measure Theory
lebesgue-measuremeasure-theory
Related Solutions
Use transfinite induction to construct $2$ (or $2^{\aleph_0}$) disjoint subsets $A_i$ of the unit interval $[0,1]$, such that each $A_i$ has nonempty intersection with every uncountable closed subset of $[0,1]$. (These are called Bernstein sets.) Then each $A_i$ has outer measure $1$, as does the union of all the $A_i$.
The following seems to be a perfect match to this question. In the following, $|*|$ denotes a Lebesgue measure, and $|*|_e$ denotes an outer measure. The outer measure does not satisfy the countable additivity on disjoint sets because of the existence of non-measurable sets. As a result, outer measure is not a measure.
Best Answer
In general what you ask cannot be true: indeed, we know that outer measure is not $\sigma$-additive but it is $\sigma$-subadditive. If it were finitely additive, it would be $\sigma$-additive.
Nevertheless, if you ask that the sets have positive distance then the answer becomes affirmative.
More precisely, if $E_1, E_2 \subset \mathbb R^n$ are such that ${\rm dist} (E_1, E_2)>0$ then $m^\star(E_1 \cup E_2) = m^\star (E_1) + m^\star(E_2)$.
This is indeed one way to show that the Lebesgue measure is a Borel measure (thanks to Caratheodory criterion).