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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.

Drew Brady
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  • 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 04 '18 at 06:48
  • 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 04 '18 at 06:49

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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 $a<b$ and $(a,b) \subset A-A^{c}$. Picking a rational number in $(a,b)$ we get a contradiction to the hypothesis.

Kavi Rama Murthy
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