Given a measurable set $E \subset \mathbb{R}^d$, with $\mathcal{H}^{d-1} (\partial E) < +\infty$, is it true in general that $E$ is a set of locally finite perimeter? that is, is it true that $\int_B |D \chi_E| dx$ is finite, for every bounded ball $B \subset \mathbb{R}^d$?

It is well-known in geometric measure theory that, in general, the perimeter $P(E)$ of a measurable set $E \subset \mathbb{R}^d$ does not equal to $\mathcal{H}^{d-1} (\partial E)$; unless $E$ has some nice regularity properties, for example when it has $C^2$ boundary. In any case, by De Giorgi's structure theorem, there is a set $\partial ^\ast E$, called the reduced boundary of $E$, which the equality $P(E)= \mathcal{H}^{d-1} (\partial^\ast E)$ holds. Recall that by definition, $P(E)<+\infty$ if the characteristic function $\chi_E$ belongs to the space $BV$ of functions with bounded variation. Thus, my question is about the existence of reduced boundary $\partial^ \ast E$ for a set $E$ with with $\mathcal{H}^{d-1} (\partial E) < +\infty$; rather than any claim about equivalence between the two boundaries. Thus it maybe true that $P(E)$ exists, but $P(E) \not = \mathcal{H}^{d-1} (\partial E)$.

It must be said that, I guess the answer is negative, but I have no idea to prove it.