# General conditions for a measurable set

Let $$A \subset [0, 1]$$ and suppose $$A$$ has finite, strictly positive Lebesgue measure. Is it possible that for all $$x, y \in A$$ it happens that $$x - y \in \mathbb{Q}$$?

My proposed solution:

Since $$A$$ has strictly positive measure, $$A$$ cannot be countable, and so $$A$$ cannot consist solely of rational numbers. Since the difference of an irrational and a rational is irrational, the answer to the question is no in this case.

Now if $$A$$ consists solely of irrationals, then I want to say that it is impossible for every difference to be rational. How do I prove this last statement?

You can use the argument about being countable again. Fix an $$x$$, then there are uncountably many differences $$x-y$$ that are all distinct. There are only countably many rationals, hence not all differences can be rational.
It is not possible. Fix $$x_0 \in A$$. If every difference is rational, then $$A \subseteq \{x_0 + q : q \in \mathbb{Q} \},$$ hence $$|A| \leq | \{x_0 + q : q \in \mathbb{Q} \} | = \sum_{q \in \mathbb{Q}} |\{x_0+q\}| = 0.$$