Show that there is a sequence of sets $A_n\subseteq [0,1]$, with outer measure 1 ($\mu^*(A_n)=1$ for every $n$), so that $A_1\supseteq A_2\supseteq A_3\supseteq...$ and $\bigcap_{n=1}^{\infty}A_n=\varnothing$.
From these conditions, the sets mustn't be measurable, so my direction is to use $\mathbb{R}/\mathbb{Q}$ in some way (like the standart construction of a non-measurable set), but I can't figure out how.
Any ideas?
The family $\mathcal F$ of nonempty open subsets of $[0,1]$ of measure $< 1$ has cardinality $c$, so it can be well-ordered so that each member of $\mathcal F$ has fewer than $c$ predecessors.
Since the complement in $[0,1]$ of any member of $\mathcal F$ has cardinality $c$, we can choose by transfinite induction, for each $U \in \mathcal F$, a countably infinite set $S(U) =\{s_1(U), s_2(U), \ldots \} \subset [0,1] \backslash U$ such that $S(U)$ is disjoint from $S(V)$ for each predecessor $V$ of $U$.
Let $$ A_n = \bigcup_{U \in \mathcal F} \{s_j(U) \; : \; j \ge n\}$$
It has outer measure $1$ because for each $U \in \mathcal F$ it contains points of $[0,1] \backslash U$. By construction, $A_{n+1} \subset A_n$. Each member of $A_1$ is $s_j(U)$ for some unique $U \in \mathcal F$ and some $j$, and thus is not in $A_n$ for $n > j$. That says $\bigcap_n A_n = \emptyset$.
I follow Cohn's Measure Theory book, in which this is featured as a problem with a very rich hint.
First a Lemma which is certainly proven in the book and which is a nice trick to have up your sleeve.
I will denote by $\mu$ the Lebesgue measure.
Lemma: If $\mu(C)>0$, there is an open interval centered at the origin such that: $$(-\varepsilon,\varepsilon)\subseteq \{x-y\:|\: x,y\in C\}$$
Proof: This follows basically from regularity of the Lebesgue measure. Measurable sets are well approximated by compact sets from the interior and open sets from the exterior. There is a compact set $K\subseteq C$ with $\mu(K)>0$. $K$ is approximated above by open sets and so there is $\mu(K)\leq \mu(U)<\mu(K)+\varepsilon'$ with $K\subseteq U$ and $U$ is open. Clearly $d=d(K,U^C)>0$, because $K$ and $U^C$ are respectively compact and closed and do not intercept. If we take $K'=K+x$ for some $x\in(-d,d)$, $K'$ will still lie within $U$ (otherwise we contradict the minimality of $d$). I claim $K$ and $K'$ share a common point, because if they were disjoint, $2\mu(K)=\mu(K)+\mu(K')\leq \mu(U)<\mu(K)+\varepsilon'$ and $\mu(K)<\varepsilon'$. But $\varepsilon'$ can be made arbitrarily small for this to be a contradiction. Thus, taking this $\varepsilon'$ at first (say $\mu(K)=\varepsilon'$) and repeating the steps, we have found $K\cap K'\not=\emptyset$ and so $k_1-k_2=x$. Thus:
$$ (-d,d)\subseteq \{x-y\:|\: x,y\in K\}\subseteq \{x-y\:|\: x,y\in C\} \tag*{$\square$} $$
With this lemma we may start attacking the problem using Cohn's hint:
Consider the Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ containing $\sqrt{p_k}$ for every prime $p_k$. This is possible because these primes are known to be linearly independent over $\mathbb{Q}$ and using Zorn's lemma we may find a maximal linearly independent set containing these square roots of prime. By taking $B$ the maximal element it is easy to see it generates all of $\mathbb{R}$ (if it didn't we could add the non generated element to the set $B$ and it would remain linearly independent, contradicting maximality of $B$).
With this we may take $B_j=B\setminus\{\sqrt{p_k}, k>j\}$. It is not hard to see $B_j\subsetneq B_{j+1}$ and $\cup_j B_j=B$. Let $S_j=\langle B_j\rangle $ be the linear subspace spanned by $B_j$. Clearly $\cup_j S_j=\mathbb{R}$ and $S_j\subsetneq S_{j+1}$. With this, we may take $A_j=[0,1]\cap S_j^C$. Now the containments are reversed and the intersection $\cap A_j =[0,1]\cap (\cup S_j)^C$ is the empty set. I have two claims:
If $\mu(C)>0$ and is contained in $S_j$, using the lemma there is $\varepsilon>0 $ such that $(-\varepsilon,\varepsilon)\subseteq \{x-y\:|\:x,y\in C\}$, we have that for any element $x\in \mathbb{R}$ there is $n$ sufficciently large for which $x/n\in (-\varepsilon,\varepsilon)$ and so $x/n=s_{j_1}-s_{j_2}$ for $s_{j_1},s_{j_2}\in S_j$, which stands in contradiction with the fact $S_j$ does not generate all sets.
Indeed, there there are open sets such that $A_j\subseteq \cup_k I_k(n)$ and $$\mu^*(A_j)\leq \mu^*(\cup I_k)\leq \sum_{k-1}^{K_n} |I_k|<\mu^*(A_j)+1/n$$
I define $U_n=\cup_k I_k(n)\cap[0,1]$. This is a sequence of open sets containing $A_j$ converging to the exterior measure of $A_j$. On Borelian sets, the exterior measure coincides with the Lebesgue measure, so:
$$\mu(U_n)+\mu([0,1]\setminus U_n])=\mu([0,1])=1$$
However, $[0,1]\setminus U_n\subseteq S_j$ as $[0,1]\cap S_j^C=A_j\subseteq U_n$. Hence, $\mu([0,1]\setminus U_n])=0$ by fact 1 and $\mu(U_n)=1$. But these were chosen so that $\mu^*(A_j)=\lim_n \mu(U_n)=1$.
