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.

Solution 1:

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).

Solution 2:

No, it isn't true. A Vitali set in [0,1] has positive outer measure. If you take the Vitali set and sufficiently many of its translates mod 1, the sum of their outer measures can exceed 1.

Specifically, let $\oplus$ denote translation mod 1. (That is, for $A \subset [0,1]$ and $x \in [0,1]$, let $A \oplus x = \{ a+x, a+x-1 : a \in A\} \cap [0,1].$) Note that outer measure $m^*$ is invariant under translation mod 1, so $m^*(A \oplus x) = m^*(A)$.

Now let $V$ be a Vitali set, and let $q_1, q_2, \dots$ be an enumeration of the rationals in $[0,1]$. By construction of $V$, the sets $V \oplus q_1, V \oplus q_2, \dots$ are pairwise disjoint and their union is $[0,1]$. By countable subadditivity we have $1 = m^*([0,1]) \le \sum_n m^*(V \oplus q_n) = \sum_n m^*(V)$; in particular we must have $m^*(V) > 0$.

So we can find an integer $N$ sufficiently large that $N \cdot m^*(V) > 1$. For $n = 1, \dots, N$, let $E_n = V \oplus q_n$. Then the sets $E_n$ are pairwise disjoint, and since $E_n \subset [0,1]$ we have $E := \bigcup_{n=1}^N E_n \subset [0,1]$. Hence by monotonicity, $m^*(E) \le m^*([0,1]) = 1$. On the other hand, $$\sum_{n=1}^N m^*(E_n) = \sum_{n=1}^N m^*(V \oplus q_n) = \sum_{n=1}^N m^*(V) = N \cdot m^*(V) > 1$$ so we have $$m^*(E) < \sum_{n=1}^N m^*(E_n).$$

Solution 3:

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.