Lebesgue measure. Find $\mu(A)$ If $I_0 = [a,b]$ and $b>a$, let $A \subset I_0$ be a measurable set such that
$$\forall p,q \in \mathbb{Q} , p \neq q \rightarrow (\{p\}+A)\cap(\{q\}+A) = \emptyset$$
Then what is $\mu(A)$?
Intuitively $\emptyset$ should imply $0$(?)
Please help!
 A: We have indeed $\mu(A) = 0.$ The reason is as follows.
Put
$$
C = \mathbb Q \cap I_0.
$$
By assumption, $C$ is countably infinite.
Further, put
$$
B = \bigcup_{r\in C} (r + A).
$$
By assumption, the union in the definition of $B$ is disjoint. Also, $B$ is measurable, being a union of measurable sets. Moreover, by translation invariance of Lebesgue measure, we have
$$
\forall r\in C: \mu(r+A) = \mu(A).
$$
Using additivity of Lebesgue measure, we can conclude
$$
\mu(B) = \sum_{r\in C}\mu(r+A) = \sum_{r\in C}\mu(A).
$$
Put
$$
J_0 = I_0 + I_0 = \{x + y | x,y \in I_0\} = [2a,2b].
$$
Observe that, since $A\subseteq I_0$ and $C\subseteq I_0,$ we have
$$
\forall r\in C: r+A \subseteq J_0,
$$
and thus also
$$
B \subseteq J_0.
$$
With all this, and using monotonicity of Lebesgue measure, we get
$$
\sum_{r\in C}\mu(A) = \mu(B) \leq \mu(J_0) = 2b - 2a < \infty.\qquad\qquad\qquad(*)
$$
If we now assume $\mu(A) > 0,$ we have a contradiction, since then the l.h.s. of (*) equals $\infty.$
So we must have $\mu(A) = 0,$ as claimed.
