Suppose that $E$ is a measurable set of real numbers with arbitrarily small periods. Explicitly, this means that there are positive numbers $p_{i}$, converging to $0$ as $i$ tends to infinity, so that
$ E + p_{i} = E$.
Prove that then eitheir $E$ or its complement has measure $0$.
Lemma: Given $A, B \subset \mathbb{R}$ of finite measure, with $\lambda(A)\lambda (B) \neq 0$, the function $x\mapsto \mu((A+x)\cap B)$ is continuous and not identically zero.
proof:
$\lambda((A+x)\cap B) = \int \chi_A(y-x)\chi_B(y)dy$ is a convolution of $L^1$ functions, hence continuous. To see its not identically zero note
$$\int \lambda((A+x)\cap B) dx = \int\int \chi_A(y-x)\chi_B(y)dydx= \int\int \chi_A(y-x)\chi_B(y)dxdy=\int \chi_B(y)\lambda(A)=\lambda(A)\lambda(B)\neq 0 $$ $\square$
Now suppose there exist $A \subset E, B \subset E^c$ with $\lambda(A)\lambda(B)\neq 0$. We may assume $A, B$ have finite measure. The set of $E$ preserving translations is a group and since it accumulates at zero, its dense. Therefore the continuous function $x\mapsto \mu((A+x)\cap B)$ is zero on a dense subset and must vanish identically. $\square \square$
