Estimate on the Hausdorff dimension of boundary of balls

Question

I am reading Evans and Gariepy's book on GMT and I have a couple questions:

1) if E is a set of locally finite perimeter, is it true that E is $ \| \partial E\|$- measurable?

2) At a certain point, he uses the estimate $ \mathcal{H}^{n-1} ( \partial B(x,r)) \leq c r^{n-1}$ where x is in the reduced boundary of E, and c is some constant. I don't know why this estimate is true.

Thank you for any help

Answer 1

1. Not necessarily. Recall that in the definition of sets of finite perimeter, as well as of the measure $\|\partial E\|$, the set $E\subset \mathbb R^n$ is identified with an element of $L^1$ represented by its characteristic function. In particular, we can add to $E$ any set of $n$-dimensional measure zero without affecting $\|\partial E\|$. If $D$ is an open disk in $\mathbb R^2$, then $\|\partial D\|$ is the linear measure on its boundary (up to a constant multiple). Adding to $D$ a non-measurable subset of $\partial D$, we obtain a set $E$ which is non-measurable with respect to $\|\partial E\|$.

2. This estimate has nothing to do with $E$ or its boundary. By the scaling property of Hausdorff measure (page 63), it suffices to prove that $\mathcal{H}^{n-1}(\partial B(0,1))$ is finite. This can be proved directly by observing that $\partial B(0,1)$ is the Lipschitz image of an $(n-1)$-dimensional cube. An alternative, and perhaps more natural way, is to appeal to surface area formulas on page 101 of the book.