Ex3-20

Question. Show that there exist disjoint E_1, E_2, \dots, E_k, \dots such that \displaystyle \left|\bigcup_k E_k\right|_e < \sum_k |E_k|_e with strict inequality. (Let E be a non-measurable subset of [0, 1] whose rational translates are disjoint. Consider the translates of E by all rational numbers r, 0 < r < 1.) 
Answer. By hint, let E be a non-measurable subset of [0, 1] whose rational translates are disjoint. Consider the translates of E by all rational numbers r, 0 < r < 1, and denote
E_r = \{x + r \in \R \mid x \in E\}.
Since 0 \le x \le 1 and 0 < r < 1, we have 0 \le x + r \le 2 so that \displaystyle \bigcup_{r \in \Q \cap (0,1)} E_r \sube [0,2] and by the monotonicty of outer measure,
\left|\bigcup_{r \in \Q \cap (0,1)} E_r \right|_e \le |[0,2]| = 2.
By Exercise 18, we have |E_r|_e = |E|_e. Moreover, |E|_e > 0 since E is non-measurable. This implies that \displaystyle \sum_{r \in \Q \cap (0,1)} |E_r|_e = \infty. Therefore we find that
\left|\bigcup_{r \in \Q\cap (0,1)} E_r \right|_e < \sum_{r \in \Q \cap (0,1)} |E_r|_e.