The Question:
Let $f\in C_c^\infty(\mathbb{R})$ be a smooth function with compact support satisfying $\int f =0$, and define $f_r(x) = r^{-1} f(x/r)$. One can show that, when $E\subset [0,1]$ is a measurable set with finite perimeter, that $$ P(E) = C \lim_{r\to 0^+} r^{-1}\|f_r\ast 1_E\|_{L^2}^2 $$ where the constant $C$ depends on the function $f$. Here $1_E$ is the indicator function of $E$.
To get intuition about this formula, observe that the convolution is zero deep in the interior or exterior of $E$, and only nonzero in a strip of width approximately $r$ on the boundary of $E$. Thus the norm captures how much mass is near the boundary, in a smoothed way.
My question is what happens when $E$ does not necessarily have finite perimeter.
Is it true that $\lim_{r\to 0^+}r^{-1}\|f_r\ast 1_E\|_{L^2}^2 = \infty$ when $P(E) = \infty$?
It turns out to be tricky, since we do not have a nice structure theorem for such sets, and in particular there does not have to exist a scale $r$ at which the behavior of the convolution "settles" into something understandable.
A Reduction
If $E\subset [0,1]$ has infinite perimeter, one can find disjoint intervals $I_1,\ldots,I_N$ in $[0,1]$ such that $0 < |I_k\cap E| < |I_k|$ for all $k$. Throwing away at most half, we can assume that the intervals are separated (none share endpoints). Since we can do this for any $N$ and using a simple rescaling argument, it suffices to prove the following statement:
Does there exist a constant $c>0$ such that the following holds:
For any measurable set $E\subset [0,1]$ with $0<|E|<1$, $$ \lim_{r\to 0^+} r^{-1}\|f_r\ast E\|_{L^2}^2 > c?$$
If we want we can even assume above that $P(E) = \infty$, though I'm not sure whether that helps. Notice that this is easy to prove when $E$ is also assumed to have finite perimeter, oddly the infinite perimeter case is what is difficult!
A Potentially useful formula
Expanding the convolution and the square, using the fact that $\int f = 0$, and applying a change of coordinates, one can show that $$ r^{-1}\|f_r\ast 1_E\|_{L^2}^2 = -\frac{1}{2r}\int |E\Delta (E+rt)| g(t)\,dt $$ where $$ g(t) = \int f(y) f(y+t)\,dy. $$ From the definition one has that $\max |g| = |g(0)| = \|f\|_{L^2}^2$, and $\int g = 0$. I can expand on the derivation of this formula if desired, though it is worth noting that the manipulations are largely the same as in this question I asked a few years ago which was motivated by the same problem.
It is also possible to express this formula using the Fourier transform, but I haven't been able to get anything from this approach that is not already contained above. Perhaps however if I had a different way to understand $\widehat{1_E}(\xi)$ I could make more progress.
Final Thoughts
While I am writing this I realize that perhaps this question is more appropriate for MathOverflow as it is a question that came up in research and I'm not sure that it has a known answer. Please let me know if this is the case. In the meantime I'd be very happy to see either a counterexample or any other angle of attack.
Thank you very much.