Average size of symmetric differences of measurable sets Let $E\subset [0,1]$ be any measurable set.  I am interested in expressions of the form
$$
I_t = \frac{1}{t} \int_0^t |E\,\Delta\, (E+s)| \,ds,
$$
and in particular of the behavior as $t\to 0$.  My question is

Are there better bounds available on $I_t$ than simply $I_t > 0$?  In particular, if $E$ is known to be a set of infinite perimeter, does it follow that $I_t = \omega(t)$?

If $E$ has finite perimeter, I already know that it is equivalent to a union of finitely many intervals, and $I_t = \Theta(t)$.  The real question is about the behavior of general measurable sets.  One can construct sets for which there is a lot of cancellation for particular choices of $s$, but since this asks about averaged information I hope that better bounds might be available.
 A: Yes, if $E$ has infinite perimeter then $I_t=\omega(t)$ as $t\to 0.$ We can write
\begin{align*}
I_t
&=\frac{1}{2t}\int_{-t}^t|E\Delta (E+s)|ds\\
&\leq\frac{1}{2t}\int_{-t}^t (1-\tfrac{|s|}t)|E\Delta (E+s)|ds\\
&=\frac{1}{2t^2}\int_{-t/2}^{t/2}\int_{-t/2}^{t/2} |E\Delta (E+s_1+s_2)|ds_1ds_2\\
&=\frac{1}{2t^2}\int_{-t/2}^{t/2}\int_{-t/2}^{t/2} |(E+s_1)\Delta (E+s_2)|ds_1ds_2\\
&=\int_{\mathbb R} d_{t/2}(x)(1-d_{t/2}(x))dx
\end{align*}
where $d_\epsilon(x)=|E\cap [x-\epsilon,x+\epsilon]|/2\epsilon$ is the $\epsilon$-approximate density of $E$ at $x,$ as featured in Lebesgue's density theorem. Note that $d_\epsilon$ is $(1/2\epsilon)$-Lipschitz.
Fix $n.$ We will show that $I_t>nt/100$ for sufficiently small $t.$ Using Lebesgue's density theorem and the infinite perimeter assumption it is easy to show that there are points $x_0<x_1<\dots<x_n$ such that $\lim_{\epsilon\to 0} d_\epsilon(x)$ exists and equals $1$ for odd $i$ and $0$ for even $i.$ This means that for sufficiently small $t,$ either $d_{t/2}(x_i)>2/3$ or $d_{t/2}(x_i)<1/3$ depending on whether $x_i$ is odd or even. By the intermediate value theorem $d_{t/2}$ must hit $1/2$ at some point $\xi_i$ between $x_i$ and $x_{i+1}$ for each $0\leq i<n.$ By the Lipschitz property $d_{t/2}(x)$ must lie in $[\tfrac13,\tfrac23]$ for $|x-\zeta_i|\leq t/6.$ So $d_{t/2}(x)(1-d_{t/2}(x))$ is at least $2/9$ over $n$ disjoint intervals each of length $t/3$; this gives $I_t\geq 2tn/27$ which suffices.
