Covering argument for Hausdorff measures So there is this proof of a lemma in  "Measure theory and fine properties of functions"  by  Evans which is on page 208. Im am confused at the part where he says "by standard covering argument we get the result" I assume that he means use vitali or besicovitch covering theorems but I don't quite see how they apply to get the desired result.. 
Note that it can be shown that $\|\partial E\|=\mathcal{H}^{n-1}|_{\partial^{*}E}$ if this is of any help.
The theorem is more or less as follows for a set $E$ with locally finite perminter we have that $\mathcal{H}^{n-1}(\partial_{*}E\setminus \partial^{*}E)=0$.
And the proof goes as follows 
Since the mapping 
$$r\mapsto \dfrac{\mathcal{L}^{n}(B(x;r)\cap E)}{r^{n}}$$
is continuous, if $x\in \partial_{*}E$ then there exists $0<\alpha<\beta<1$ and $r_{j}\to 0$ such that 
$$\alpha\leq \dfrac{\mathcal{L}^{n}(B(x;r_{j})\cap E)}{\omega_{n}r_{j}^{n}}\leq \beta$$
Thus 
$$\min\{\mathcal{L}^{n}(B(x;r_{j})\cap E),\mathcal{L}^{n}(B(x;r_{j})\setminus E)\}\geq \min\{\alpha,1-\beta\}\omega_{n}r^{n}_{j}$$
so the Isoperimetric inequality gives us that 
$$\limsup_{r\to 0}\dfrac{\|\partial E\|(B(x;r))}{r^{n-1}}>0.$$
Since $\|\partial E\|(\mathbb{R}^{n}\setminus \partial^{*}E)=0$ standard covering argument imply $\mathcal{H}^{n-1}(\partial_{*}E\setminus \partial E^{*})=0$
 A: I had the exact same question when I was reading this section recently. I'm not sure if the following is the "standard covering argument", but here is my attempt (subject to possible errors!)
For $x \in \partial_*E$, let $c(x): =\limsup_{r \to 0}\frac{\|\partial E\|(B(x,r))}{r^{n-1}} > 0$. Unfortunately, this is a pointwise lower bound, and we ideally want a uniform bound to use Vitali's Covering Theorem. 
To fix this, let $S = \partial_*E\backslash \partial^* E$, and define
\begin{equation}
L_k = \{ x \in S\,:\, c(x) > \frac{1}{2^k}\}.
\end{equation}
We have $S = \bigcup_{k = 1}^{\infty} L_k$, and $\|\partial E\|(L_k) = 0$. (The measurability of the $L_k$'s is not a concern as we are dealing with outer measures here.)
Let $\epsilon > 0$. As $\|\partial E\|$ is a Radon measure, there is an open $U_k \subset \mathbb{R}^n$ containing $L_k$ such that 
\begin{equation}
\|\partial E\|(U_k) \leq \|\partial E\|(L_k) + \epsilon2^{-k} = \epsilon2^{-k}
\end{equation}
Let $\delta > 0$, and define
\begin{equation}
 \mathcal{F}_k := \left\{ B(x,r)\,:\, x \in L_k, \,B(x,r) \subset U_k,\, 0 < r < \frac{\delta}{10}, \,\|\partial E\|(B(x,r)) \geq \frac{r^{n-1}}{2^k}\right\}
\end{equation}
$\mathcal{F}_k$ is a covering of $L_k$, so by Vitali, there exists a countable disjoint collection $\{B(x_i, r_i)\}_{i = 1}^{\infty} \subset \mathcal{F}_k$ such that
\begin{equation}
L_k \subset \bigcup_{i = 1}^{\infty}B(x_i, 5r_i).
\end{equation}
As $\text{diam} B(x_i, 5r_i) < \delta$, this collection is admissible in the definition of $\mathcal{H}_{\delta}^{n-1}(L_k)$, so 
\begin{align}
\mathcal{H}_{\delta}^{n-1}(L_k) &\leq \omega_{n-1}\sum_{i = 1}^{\infty}(5r_i)^{n-1}\\
&\leq \omega_{n-1}5^{n-1}2^k \sum_{i = 1}^{\infty}\|\partial E\|(B(x_i,r_i))\\
&= \omega_{n-1}5^{n-1}2^k\|\partial E\|(\bigcup_{i = 1}^{\infty} B(x_i,r_i))\\
&\leq \omega_{n-1}5^{n-1}2^k\|\partial E\|(U_k)\\
& \leq \omega_{n-1}5^{n-1}\epsilon.
\end{align}
Letting $\epsilon \to 0$, we have $\mathcal{H}_{\delta}^{n-1}(L_k) = 0$, and therefore
\begin{equation}
\mathcal{H}_{\delta}^{n-1}(S) = \mathcal{H}_{\delta}^{n-1}(\bigcup_{k = 1}^{\infty}L_k) = \lim_{k \to \infty}\mathcal{H}_{\delta}^{n-1}(L_k) = 0.
\end{equation}
Hence $\mathcal{H}^{n-1}(S) = 0$.
