Suppose $s:[0,1]\to\mathbb R$ is a simple function with $s(0)=s(1)$. Let $S:\mathbb R \to \mathbb R$ be $S(x)=s(x-\lfloor x\rfloor)$, the function that repeats $s$ on each interval $[n,n+1]$. I'm trying to decide if for all $\epsilon>0$ there is a $\delta>0$ so that whenever $|r|<\delta$ we have $\int_0^1|S(t)-S(t+r)|\ dt<\epsilon.$ The integral here is the Lebesgue integral.
Any help is very appreciated. Perhaps it's not true but I've been unable to find a counter example.
Here is an outline of the proof.
1) Prove it when $s$ is the indicator of an open interval $I \subset (0,1)$. This is a straightforward computation.
2) Prove it when $s$ is the indicator of an open set $U \subset (0,1)$. An open set of finite measure is almost a finite union of disjoint open intervals.
3) Prove it when $s$ is the indicator of a measurable set $E \subset (0,1)$. Any measurable set is almost an open set.
4) Prove it for simple functions $s$ - this should just be an application of the triangle inequality.