I saw an answer here on the page stating the fact that for a dense sequence $(a_n)$ in $[0, 1]$ and a set of positive measure $A$, the set $\bigcup_{n \geq 1} A + a_n$ has measure $1$ (with respect to the Lebesgue-measure). So I tried to prove it.

Let $B := \left( \bigcup_{n \geq 1} A + a_n \right) \cap \mathbf{R}/ \mathbf{Z}$.

My attempt was to pick an open set $U \supset B$ with $\lambda(U) \leq \lambda (B) + \varepsilon$ for an arbitrary $\varepsilon >0$. This can be done since the Lebesgue measure is outer regular.

But then $U$ contains a dense sequence, hence any non-empty open set $V$ intersects it (the set $U$ being non-empty, as $A$ has positive measure). But then again, as $[0, 1]$ is Hausdorff, we can only have $U = [0, 1]$. As $\varepsilon$ was arbitrary, the set $B$ has measure $1$.

However I'm a bit confused about my "proof". If $B$ were $\mathbf{Q} \cap [0, 1]$, the same argument would show that the measure of $\mathbf{Q} \cap [0, 1]$ is one.

Where is my mistake?

And also: How can one prove the statement?


  • $\begingroup$ If anyone is interested in the question where I found the fact, here's the link: math.stackexchange.com/questions/1033251/… $\endgroup$ – Steven Jun 2 '17 at 16:36
  • $\begingroup$ Do you mean $(\cup (A+a_n)) \cap [0,1]?$ $\endgroup$ – zhw. Jun 2 '17 at 16:59
  • $\begingroup$ yes, of course, I'll edit it $\endgroup$ – Steven Jun 2 '17 at 17:03
  • $\begingroup$ Perhaps you mean that there exists a translation of $\bigcup_n A+a_n$ which has full measure in $[0,1]$. $\endgroup$ – Rigel Jun 2 '17 at 17:10
  • $\begingroup$ I mean the elements seen to be in $[0, 1) $ modulo $1$. so yes, you're right. $\endgroup$ – Steven Jun 2 '17 at 17:17

Let us prove that there exists a translation of the set $\bigcup_n (A + a_n)$ which has full measure in $[0,1]$.

Assume by contradiction that $B :=\bigcup_n (A + a_n)$ has not full measure in $[0,1]$, so that there exists a point $x\in B$ with density $0$ in $B$. Let us fix $\epsilon \in (0,1/3)$. Hence there exists $\rho_0 > 0$ such that $$ \frac{\lambda((x-\rho, x+\rho)\cap B)}{2\rho} < \epsilon \qquad \forall \rho \in (0, \rho_0). $$ To simplify the argument, we also assume that $(x-\rho_0, x+\rho_0) \subset [0,1]$.

W.l.o.g. we can assume that $0$ is a Lebesgue point of $A$, i.e., there exists $r \in (0, \rho_0)$ such that $$ \frac{\lambda(A_r)}{2r} > 1 - \epsilon, \quad\text{where}\quad A_r := ( - r, r) \cap A. $$ It is not restrictive to assume $r < 2\epsilon\rho_0$.

On the other hand, $B \supset C_r := \bigcup_n (A_r + a_n)$.

Since $(a_n)$ is dense in $[0,1]$, given $\rho :=r/{2\epsilon} \in (r, \rho_0)$ there exists an index $j\in\mathbb{N}$ such that $A_r + a_j \subset (x-\rho, x+\rho)$, hence $$ A_r + a_j \subset (A_r + a_j) \cap B \subset (x-\rho, x+\rho)\cap B $$ so that $$ 2 r (1-\epsilon) < 2\rho \epsilon = r, $$ a contradiction.

  • $\begingroup$ I don't quite see why the last inequality holds. Could you maybe elaborate on that? $\endgroup$ – Steven Jun 2 '17 at 17:30
  • $\begingroup$ I have modified the proof. (Not sure of the choice of the constants, but the idea should be correct.) $\endgroup$ – Rigel Jun 2 '17 at 18:04

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