Lebesgue measure. Find $\mu(A)$

Question

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!

Answer 1

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.