Invariant property in Lebesgue measure

Question

Let $E \subset \mathbb{R}$ be a Lebesgue measruable set.

I proved that

\begin{align*} \lim_{t\rightarrow 0} m(E\cap(E+t))=m(E) \end{align*}

by applying Dominated Convergence theorem to $\chi_{E}(x)\chi_{E}(x-t) $.

I want to know that

Let $E \subset [0,1]$ be a Lebesgue measurable set such that \begin{align*} m(E\cap(E+t))=(m(E))^2, \qquad \textrm{for all} \quad \vert t\vert\leq 1/2. \end{align*} What is $m(E)$?

By previous problem, there exists $\delta >0$ such that $m(E\cap (E+t))=m(E)$ if $\vert t \vert <\delta$. Thus, in $\vert t \vert <\delta$, we get $m(E)=(m(E))^2$. I thought that $m(E)= 0$ or $1$. However, I couldn't choose whether $m(E)$ is 0 or 1..

Any help is appreciated..

Thank you!

Answer 1

If $\mu(E) =1$, then $\mu([0,1] \setminus E) =0$. Thus $\mu([0,1] \cap (E+t)) =1$ for all $|t|\leq 1/2$. Using translations invariance, we get $\mu(([0,1]+t) \cap E) =1$ for all $|t| \leq 1/2$.

Taking $t=1/2$ and $t=-1/2$, then $\mu([-1/2,1/2] \cap E) =1$ and $\mu([1/2,3/2] \cap E) =1$. Thus $\mu(E) \geq 2$. A contradiction!

That's exactly your argument in the comments - as I see now.