Outer measure does not satisfy countable additivity

Question

Let $\Omega \subseteq \Bbb R^n$, we define the outer measure the following way : $m^*(\Omega):= \inf\{ \sum_{j \in J} \text{vol } B_j \lvert \{B_j\} \text{ is countable covering by boxes of } \Omega \}$ where a box $B$ is $B:=\Pi_{i=1}^n (a_i,b_i)$ and vol($B)=\Pi_{i=1}^n (b_i-a_i)$.

It is said in my lectures notes that this outer measure does not satisfy countable additivity : if $A,B$ disjoint then $m^*(A \cup B) = m^*(A) + m^*(B)$.

Can you give me a counterexample for this i.e. $A,B$ disjoint sets such that $m^*(A \cup B) \neq m^*(A) + m^*(B)$.

Answer 1

$\newcommand{\a}{\mathcal{A}}\newcommand{\p}{\mathcal{P}}\newcommand{\i}{\mathcal{I}}$I will demonstrate a favourite proof of mine.

Theorem: there does not exist any set function $\mu:\p(\Bbb R)\to[0,\infty]$ that satisfies:

1. $$\mu\left(\bigsqcup_{n\in\Bbb N}E_n\right)=\sum_{n\in\Bbb N}\mu(E_n)$$
2. $$\mu(E+t)=\mu(E),\,\forall t\in\Bbb Q$$
3. $$0\lt\mu((0,1]]\lt\infty$$

Note that $\sqcup$ denotes disjoint union. Since $m^\ast$ satisfies $(2),(3)$ and you want to show it does not satisfy $(1)$, you can let $\mu=m^\ast$ to get the same result. More generally, condition $(3)$ can be relaxed to say: there is at least one open set $G\in\p(\Bbb R)$ with $0\lt\mu(G)\lt\infty$.

Proof:

Suppose such a $\mu$ exists. Let $0\lt\mu((0,1])=\alpha\lt\infty$.

Let $x\simeq y\iff x-y\in\Bbb Q$. This is a proper equivalence relation, so we can let $\i=(0,1]_{/\simeq}$. By the Axiom of Choice, there is a choice function $f:\i\to(0,1]$. Let: $$\a=\{f(x):x\in\i\}$$By the countability of the rationals, we can find an enumeration $\{r_k:k\in\Bbb N\}$ of $(-1,2]\cap\Bbb Q$. Let $\a_k:=\a+r_k$. All the $\a_k$ are disjoint; suppose otherwise, that there are $m,n\in\Bbb N$ and an $x\in\a_m\cap\a_n$. We can say $x=a_m+r_m=a_n+r_n$ where $a_m,a_n\in\a$: then $a_m-a_n=r_n-r_m\in\Bbb Q$, so we must have that $a_m\simeq a_n$ and $a_m=a_n$ by construction of $\a$. Then $r_m=r_n$ and $n=m$ since $\{r_k\}$ is an enumeration, hence $\a_m=\a_n$: therefore $\{\a_k\}$ is a family of disjoint sets. They are also all rational translations of the same set $\a$, so by $(2)$ we have $\mu(\a_k)=\mu(\a)$ for all $k$.

Observe that: $$(0,1]\subseteq\bigsqcup_{k\in\Bbb N}\a_k\subseteq(-1,2]=(-1,0]\sqcup(0,1]\sqcup(1,2]$$Then by $(1)$ and $(2)$ we have: $$\alpha\le\sum_{k\in\Bbb N}\mu(\a_k)=\sum_{k\in\Bbb N}\mu(\a)\le3\alpha$$Since $\alpha\gt0$ by assumption $(3)$, there exists no real number $\mu(\a)$ such that this inequality holds. Hence $\mu(\a)$ cannot exist in $\Bbb R$, a contradiction to $\mu$ being defined for every subset of $\Bbb R$. $\blacksquare$

Again, everywhere you could have replaced $\mu$ with $m^\ast$ and shown that $m^\ast$ cannot satisfy countable additivity - condition $(1)$ - on the pathological family $\{\a_k:k\in\Bbb N\}$.

For Lebesgue's measure, pathological sets like $\a$, which "change" in (outer) measure as you move them, are deemed non-measurable; we restrict the sigma algebra to a certain collection of nice sets - $\a$ is definitely not in that collection. This is just one of many examples.

Answer 2

You need to look for weird sets, like a Vitali set.