Let $A,B \subset S^1$ be measurable sets (considering $S^1$ with say the lebesgue measure). I'm trying to prove that if the convolution $1_A*1_B$ is constant then one of $A$ or $B$ is a full measure set or one of them is of measure zero.
I didn't make much progress but i'm pretty sure that the measure of the intersection of two sets can't be invariant to translation of one of the sets. It's just too pathological. Thank you for the help.
Take $A=[0,\pi]$ and $B=[0,\pi/2]\cup [\pi,3\pi/2]$. Then $I_A*I_B(y)=\mu((y-A)\cap B)\equiv const$. It can either be seen geometrically, or checked formally by either writing the explicit formula or caculating the Fourier coefficients $\widetilde{I_A*I_B}=\tilde{I_A}\tilde{I_B}$ and noticing that $\tilde{I_A}(2n)=0=\tilde{I_B}(2n-1)$ for $n\ge 1$, so the only non-zero coefficient in the Fourier series (in $L_2$ sense) is the one before the constant. The last argument represents the idea on how such sets can be constructed.