$m(E)=0$ or $m(E^c)=0$ The question comes from former qualifying exam of the graduate school I'm going to attend.
Q: Suppose $E$ is measurable and $E=E+\frac{1}{n}$ for every natural number $n\geq 1$. Show that either $m(E)=0$ or $m(E^c)=0$.
So far all I've got is by induction $E=E+q$ for $q\in Q$. That is, if we define the equivalence relation on $R$ by $x$ ~ $y \iff x-y\in Q$ then $E$ either contains all elements or no element in each equivalent class. I'm not sure how this help though.
 A: If $I$ is an interval say $$A(I)=\frac{m(I\cap E)}{m(I)}.$$
Say $D_n$ is the collection of dyadic intervals of length $2^{-n}$, which is to say the elements of $D_n$ have the form $[j2^{-n},(j+1)2^{-n})$ for some integer $j$.
What you've shown shows that $$A(I)=A(J)\quad(I,J\in D_n).$$
Now say $I_L$ and $I_R$ are the left and right halves of $I\in D_n$. Then $A(I_L)=A(I_J)$, and a moment's thought shows that $$A(I)=\frac12(A(I_L)+A(I_R)),$$so $A(I_L)=A(I)$. And so $$A(I)=A(J)\quad(I,J\in\bigcup_n D_n).$$Say $A(I)=\alpha=A([0,1])$ for every dyadic interval $I$. It's enough to show that $\alpha=1$ if $m(E)>0$. 
Now for every $x$ there exists $I_n(x)\in D_n$ with $x\in I_n$. The Lebesgue differentiation theorem shows that $$\lim_{n\to\infty}A(I_n(x))=1$$for almost every $x\in E$. In particular this happens for at least one $x$ if $m(E)>0$, showing that $\alpha=1$ in that case.
A: Two bonus solutions: First a much simpler version of the argument in the other answer, and second a totally different proof by Fourier series.
First Let $f(x)=m(E\cap[0,x])$. Then $f$ is continuous, and the Lebesgue differentiation theorem shows that it is differentiable almost everywhere and that the derivative is $0$ or $1$ almost everywhere.
But the hypothesis implies that $$f(x+1/n)-f(x)=f(1/n)-f(0).$$
So given that $f$ is differentiable almost everywhere it is actually differentiable everywhere, and in fact $f'(x)=f'(0)$. If $f'=0$ then $f=0$ and so $m(E)=0$, while if $f'=1$ then $f(x)=x$, showing that $E$ has full measure.
Second Let $f=\chi_E$. Regarding $f$ as a function with period $1$ it has Fourier coefficients $$\hat f(k)=\int_0^1f(t)e^{-2\pi i kt}\,dt\quad(k\in\Bbb Z).$$Since $f$ has period $1$ you can take the integral over any interval of length $1$. So $$\hat f(k)=\int_0^1f(t+1/n)e^{-2\pi i kt}\,dt=
\int_{1/n}^{1+1/n}f(t)e^{-2\pi ik(t-1/n)}\,dt=e^{2\pi ik/n}\hat f(k).$$
This shows that $\hat f(k)=0$ for $k\ne0$, so $f$ is (essentially) constant.
