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 $1$ 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?
Thanks!
 A: 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.
