You are given an interval $[a,b]$ (you can assume WLOG that $a<b$) and you take $A \subset [a,b]$ as a measurable set such that: $$\forall_{c,d\in Q} c\neq d \rightarrow (\{c\}+A) \cap (\{d\}+A))= \emptyset $$
Now show the measure of the set, $\mu(A)$ is equal to $0$.
I am confused because to me this implies that the interval is empty, but clearly that's not true because we're told to assume that a is less than b.
I'm assuming that $\{c\}+A$ denotes the set $\{c+a:a\in A\}$, and that $\mu$ denotes Lebesgue measure.
For $n \in \mathbb{N}$, define $$A_n = \bigcup_{k \in \mathbb{Z}} \{-k/n\}+\big(A \cap\big[k/n,(k+1)/n\big)\big)$$ From the given condition, this union is disjoint. Hence, by translational invariance of $\mu$, $$\mu(A_n) = \sum_k \mu\big(\{-k/n\}+A \cap[k/n,(k+1)/n)\big) = \sum_k \mu\big(A \cap[k/n,(k+1)/n)\big) = \mu(A)$$
It is easily checked that $A_n \subset [0,\frac{1}{n})$ for all $n \in \mathbb{N}$ (since each term in the union is contained in $[0,\frac{1}{n})$).
Since $A_n \subset [0,\frac{1}{n})$ it follows that $\mu(A) = \mu(A_n) \leq \frac{1}{n}$ for all $n$, and thus $\mu(A)=0$.