8
$\begingroup$

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 $0.6 + 0.5 = 0.1$.

$\endgroup$
3
$\begingroup$

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.

Note that a cyclic shift is a homeomorphism of the unit interval. Thus,

$$ \bigcap_i K+x_i = \bigcap_{i, n} (U_n +x_i) $$ Is uncountable (as seen above) since it is a countable intersection of open dense sets.

$\endgroup$
  • $\begingroup$ 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 $1 - \frac1n$. $\endgroup$ – Ilya Nov 13 '15 at 7:33
  • $\begingroup$ @Ilya: I expanded my answer a bit. $\endgroup$ – PhoemueX Nov 13 '15 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.