# Not a small, not a big set

Consider a unit interval $X = [0, 1]$ endowed with Borel $\sigma$-algebra generated by the usual topology. I am looking for a measurable set $K\subseteq X$ which has a $0$ Lebesgue measure, but such that $\bigcap_{i}(K + x_i)$ is uncountable for any countable sequence of $x_i\in X$. Here the addition is cyclical, that is I assume $0.6 + 0.5 = 0.1$.

Note that the Baire category theorem implies that if $U_n$, $n\in \Bbb {N}$ are open dense sets, then $\bigcap U_n$ is dense. In particular, the intersection is uncountable, since if it was of the form $\{x_n \mid n\}$, we could set $V_n = [0,1]\setminus \{x_n\}$ (which is open and dense), so that Baire again implies that $\bigcap U_n \cap V_n =\emptyset$ is dense, which is absurd.
Now, let $K =\bigcap U_n$ be a null set, where the $U_n$ are open dense sets. Existence of such a set can be seen as follows: Let $Q \subset [0,1]$ be countable, dense, for example $Q = \Bbb{Q} \cap [0,1]$. Then the Lebesgue measure of $Q$ is $\lambda(Q) = 0$. Thus, by outer regularity, there is for each $n \in \Bbb{N}$ an open set $U_n \supset Q$ (in particular, $U_n$ is dense in $[0,1]$) with $\lambda(U_n) < 1/n$. Then clearly, $K = \bigcap_n U_n$ satisfies $0 \leq \lambda(K) \leq \lambda(U_n) <1/n$ for all $n$, so that $K$ is a null set as desired.
$$\bigcap_i K+x_i = \bigcap_{i, n} (U_n +x_i)$$ Is uncountable (as seen above) since it is a countable intersection of open dense sets.
• You mean $U_n$ are open dense sets? Note sure how to apply outer regularity to show existence, but I guess I can just take $U_n = [0,1]\setminus C_n$ where $C_n$ is a fat Cantor set of measure $1 - \frac1n$. – Ilya Nov 13 '15 at 7:33