I read this statement in the book by Evans & Gariepy, page 215, last two lines.
Here $f\in BV(R^n)$ and for fixed $\epsilon>0$ and $N>0$, we define $$A_\epsilon^N :=\{x\in \mathbb R^N,\,\,\limsup_{r\to 0}\frac{\|D(f-M)^+\|( B(x,r))}{r^{n-1}}>\epsilon\text{ for all }M\geq N\} $$
Then the book claims, without proof, that $$\epsilon\mathcal{H}^{n-1}(A_\epsilon^N)\leq C\|D(f-M)^+\|(R^n)=C\int_M^\infty\|\partial E_t\|(\mathbb R^n)dt$$
I understand how second equality come from, but I got confused on the first inequality. It looks to me that the book somehow use Chebyshev's Inequality but I don't know how to write done the details.
Fix $\delta>0$ such that the Hausdorff pre-measure $\mathcal H_\delta^{n-1}(A_\epsilon^N)$ is at least $\frac12 \mathcal H ^{n-1}(A_\epsilon^N)$.
For each $x\in A_\epsilon^N$, pick $r_x<\delta/2$ so that
$$\|D(f-M)^+\|( B(x,r_x))>\epsilon r_x^{n-1}$$ The balls $ B(x,r_x)$ form a Besicovitch cover of $A_\epsilon^N$; the Besicovitch covering theorem provides several disjoint subcollections $B_k$, $k=1,\dots,\mathcal N$ of these balls which together cover $A_\epsilon^N$. We have
$$\sum_{k=1}^{\mathcal N}\sum_{B(x,r_x)\in B_k} r_x^{n-1}\ge c \mathcal H_\delta^{n-1}(A_\epsilon^N) $$ by the definition of the right hand side. Also, $$\begin{split}\sum_{k=1}^{\mathcal N} \sum_{B(x,r_x)\in B_k} r_x^{n-1} &\le \epsilon^{-1} \sum_{k=1}^{\mathcal N}\sum_{B(x,r_x)\in B_k}\|D(f-M)^+\|( B(x,r_x)) \\ & \le \epsilon^{-1} \mathcal{N}\|D(f-M)^+\|(\mathbb R^n) \end{split} $$