Measure of a set invariant under translations with rational numbers Suppose $E\subset\mathbb{R}$ is measurable and $E=E+\frac{1}{n}$ for every natural number $n\geq1$. Show that either $m(E)=0$ or $m(\mathbb{R}\setminus E)=0$. 
 A: Let $m(E)>0,a\in 
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\mathbb{R}
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$ and $f(x)=m(E\cap \lbrack a,x])$ for $a\leq x<\infty $. If $a<x<y$ then $$
f(y+\frac{1}{n})-f(x+\frac{1}{n})=m(E\cap (x+\frac{1}{n},y+\frac{1}{n}])$$
$$=m(\{E-\frac{1}{n}\}\cap (x,y])=m(E\cap (x,y])$$ and $$f(y-\frac{1}{n})-f(x-%
\frac{1}{n})=m(E\cap (x-\frac{1}{n},y-\frac{1}{n}])$$
$=m(\{E+\frac{1}{n}\}\cap (x,y])=m(E\cap (x,y])$. It follows that $$f(y+\frac{%
1}{n})-f(x+\frac{1}{n})=f(y-\frac{1}{n})-f(x-\frac{1}{n})$$ . Note that $$%
\left\vert f(y)-f(x)\right\vert \leq \left\vert y-x\right\vert $$ so $f$ is
absolutely continuous. Hence it is differentiable almost everywhere and
using above equation we conclude that its derivative is a constant $c$ a.e..
Since $$\frac{m(x-\delta ,x+\delta )}{2\delta }=\frac{f(x+\delta )-f(x-\delta
)}{2\delta }$$ and almost all points of $E$ have metric density $1$ we see
that $c=1$ Thus $f(y)-f(x)=\int_{x}^{y}f^{\prime }(t)dt=y-x$. This
gives $f(y)=f(a)+y-a$ $\forall x>a$. Thus $f(y)-f(x)=y-x$ or $m(E\cap
(x,y])=m((x,y])$ for $a<x<y$. This gives $m(E^{c}\cap (x,y])=0$ for $a<x<y$
which clearly implies that $m(E^{c})=0$.
