# Hint on measure theory problem about rational-invariant measurable sets

I'm trying to solve the following problem:

Suppose that $$A \subset \mathbf{R}$$ is Lebesgue measurable and is such that for each $$x \in A$$, $$x + \mathbf{Q} \subset A$$. Show that $$\lambda(A)$$ or $$\lambda(\mathbf{R} \setminus A)$$ is 0.

Above, $$\lambda$$ denotes the Lebesgue measure on $$\mathbf{R}$$.

One thing I noticed is that with $$B = A \cap [0, 1)$$, we have $$A = \cup_{n \in \mathbf{Z}} (B + n)$$, which is a disjoint union, and thus by countable additivity and translation invariance it suffices to show that $$\lambda(B) \in \{0, 1\}$$. Since $$B$$ is Lebesgue measurable,
$$f(x) := \lim_{\epsilon \to 0} \frac{\lambda(B \cap (x - \epsilon, x + \epsilon)) }{2\epsilon}$$ is such that $$f(x) = 1$$ for $$\lambda$$-a.e. $$x \in B$$. I don't really know where to go from here.

• Hint: First note that it suffices to show $A \cap [0,1]$ has measure 1 or 0. Show that if $A \cap [0,1]$ has measure $\epsilon$ then show $A \cap (a,b)$ has measure $\epsilon (b - a)$ when $b$ and $a$ are rational. Show that this cannot happen, unless $\epsilon = 1$. – Sean Haight Dec 4 '18 at 6:48
• Yes, I agree it clearly suffices to show that $A \cap [0, 1)$ is zero or 1 with respect to Lebesgue measure, due to additivity and translation invariance. – Drew Brady Dec 4 '18 at 6:49

It is known that if $$\lambda (E) >0$$ and $$\lambda (F) >0$$ then $$E+F$$ contains an open interval. (You can find a proof on MSE: "Sum" of positive measure set contains an open interval?). Assume that $$\lambda (A) >0$$ and $$\lambda (A^{c}) >0$$ ($$A^{c}=\mathbb R\setminus A$$). Then there exist $$a,b$$ with $$a and $$(a,b) \subset A-A^{c}$$. Picking a rational number in $$(a,b)$$ we get a contradiction to the hypothesis.