$m^\star(A\cup B)<m^\star(A)+m^\star(B)$

Question

Show that there are disjoint sets $A$ and $B$ such that $m^\star(A\cup B)<m^\star(A)+m^\star(B)$

I know how to prove,

If $A$ and $B$ be bounded subsets of $\mathbb{R}$ for which $d(A,B)>0$ then $m^{*}(A\cup B)=m^*(A)+m^*(B)$.

By constructing open set, $$U=\bigcup_{x\in A}\left(x-\frac{\alpha}2,x+\frac{\alpha}2\right)$$ And using $m^*(A \cup B) = m^*((A \cup B) \cap U) + m^*((A \cup B) \cap U^c)$.

But for $m^\star(A\cup B)<m^\star(A)+m^\star(B)$, I couldn't think of any idea to construct such set $A,B$.

Any help will be appreciated. TIA

Answer 1

Let denote operation $\oplus$ on $[0, 1)$: $x \oplus y = (x + y) \pmod 1$. Note that $\mu^*(A \oplus x) = \mu^*(A)$ for any $A \subset[0, 1)$ and any $x$.

Let $X$ be Vitali set on $[0, 1)$ (such set that $X \oplus q$ are pairwise disjoint for rational $\mathbb Q \cap [0, 1)$ and $\cup_q X \oplus q = [0, 1)$).

If $\mu^*(X) = 0$ then $\mu^*([0, 1)) = \mu^*(\cup_q (X \oplus q)) \leq \sum_q \mu^*(X \oplus q) = 0$. So, $\mu^*(X) > \frac{1}{n}$ for some $n$.

Now, choose $n+1$ rational numbers from $[0, 1)$ - $q_1, q_2, \ldots, q_{n + 1}$.

We have $\mu^*((X \oplus q_1) \cup (X \oplus q_2) \cup \ldots \cup (X \oplus q_{n + 1})) < \frac{n + 1}{n}$, as in the left part we have measure of subset of $[0, 1)$.

Let $m$ be minimal number s.t. $\mu^*((X \oplus q_1) \cup \ldots \cup (X \oplus q_m)) < \frac{m}{n}$. Let $A = (X \oplus q_1) \cup \ldots (X \oplus q_{m - 1})$ and $B = X \oplus q_m$.

Then $\mu^*(A) = \frac{m - 1}{n}$, $\mu^*(B) = \frac{1}{n}$, but $\mu^*(A \cup B) < \frac{m}{n}$.

Alternatively, if measurable set is defined as set that has equal inner and outer measure, take $A$ as any non-measurable subset of $[0, 1]$, $B = [0, 1] \setminus A$, and note that as $A$ is not measurable, $\mu^*(A) + \mu^*(B) = \mu^*(A) + (1 - \mu_*(A)) = 1 + (\mu^*(A) - \mu_*(A)) > 1 = \mu^*([0, 1]) = \mu^*(A \cup B)$.