Originally Posted by

**Opalg** 1) Given m, n and k, the box

has volume 2/k. Taking the intersection over all natural numbers k, you see that the set

has measure 0. Now take the (countable) union over all integers m and n.

2) By the same argument as in 1), the set

has measure 0, for each rational number r. Now take the union over all such r.

Opalg's eplanation is clear. If I understand Opalg's 1), then by parrallel argument, should 2) be unioned over three indexes?:

2) By the same argument as in 1), the set

has measure 0, for any rational number r. Now take the (countable) union over all integers m, n and p.