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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?

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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.

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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 $$|A| \leq | \{x_0 + q : q \in \mathbb{Q} \} | = \sum_{q \in \mathbb{Q}} |\{x_0+q\}| = 0.$$

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