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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.

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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 $$m^*(E) < \sum_{n=1}^N m^*(E_n).$$

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  • $\begingroup$ Thank you very much but your last sentence "the sum of their outer measures can exceed 1" is correct, but didn't answer my question. I don't know how to find the outer measure of a union of finitely many translates of a Vitali set. $\endgroup$ – Singfook Sangwood Jul 26 '14 at 19:47
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    $\begingroup$ @user160307: The union of the translates is still a subset of [0,1]. We don't need to compute its outer measure exactly; by monotonicity it is certainly at most 1. $\endgroup$ – Nate Eldredge Jul 26 '14 at 20:53
  • $\begingroup$ But how sum of outer measure of finitely many translates of Vitali set(chosen properly) can extend 1.(We at least have to give proper counterexample). $\endgroup$ – Sushil Sep 13 '14 at 11:35
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    $\begingroup$ @Sushil: I don't completely understand your objection, but I have added more details, so hopefully it is clear now. $\endgroup$ – Nate Eldredge Sep 13 '14 at 15:01
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    $\begingroup$ @NateEldredge I don't had any objection the only problem was I didn't get the argument which is clear now. Thanks for editing $\endgroup$ – Sushil Sep 13 '14 at 16:03
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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).

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  • $\begingroup$ Thanks. I am a beginner in measure theory and I could have found that the original statement was trivially false. But your condition such that the sets have positive distance is quite interesting. $\endgroup$ – Singfook Sangwood Jul 28 '14 at 18:11
  • $\begingroup$ You are welcome. $\endgroup$ – Romeo Jul 28 '14 at 20:40
  • $\begingroup$ @Romeo Is any outer measure finitely additive on the sets of positive distance ? $\endgroup$ – W. Volante Jan 14 '18 at 21:11
  • $\begingroup$ @W.Volante I do not think so. Take $X=\{1,2,3\}$ with the induced euclidean metric. Set $\mu(\emptyset)=0$, $\mu(X)=2$ and $\mu(E)=1$ for any other subset of $X$. This $\mu$ is an outer measure but $\mu(\{1\} \cup \{2\}) = 1$ while $\mu(\{1\})+\mu(\{2\})=2$. Does this answer to you? $\endgroup$ – Romeo Jan 17 '18 at 23:32
  • $\begingroup$ @Romeo yes, thanks you. $\endgroup$ – W. Volante Jan 18 '18 at 19:46
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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.

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  • $\begingroup$ You have written: If we assume that m∗([0,1]∩X)<1 then there will be a non–empty perfect subset Y⊂[0,1] such that Y∩([0,1]∩X)=∅. Why so? $\endgroup$ – Sushil Sep 13 '14 at 11:31
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I found this proof in Real Analysis by Royden: Theorem: There are disjoint sets of real numbers A and B for which m*(A U B) < m*(A) + m*(B). Proof We prove this by contradiction. Assume m*(AU B) = m*(A) + m*(B) for every disjoint pair of sets A and B. Then, by the very definition of measurable set, every set must be measurable. This contradicts the existence of a non-measurable set.

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It is a theorem of Lusin that the restriction of Lebesgue outer measure $\mu^*$ to any non null set X cannot be finitely additive. See the first theorem here: http://www.math.wisc.edu/~akumar/Thesis.pdf

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