I want to show that the mapping $\mathbb{R}\ni t\mapsto \mathcal{H}^{n-1}(\Omega \cap \{x:u(x)>t\}) = \sup\left\{\displaystyle\int_{\{x:u(x)>t\}}div(v)\,dx, v=(v_1,...,v_n)\in C^{\infty}_0(\Omega, \mathbb{R}^n), \|v\|_\infty\leq 1 \right\}$
is measurable. Here $\Omega\subset\mathbb{R}^n$ is open and $u$ is a BV function. I have tried it with the usual argument, having a look at $\left(\mathcal{H}^{n-1}(\Omega \cap \{x:u(x)>t\}\right)^{-1}((0,a])$ for $a\in\mathbb{R}$ arbitrary but this didn't lead to anything as I had problems dealing with the Hausdorff-measure. Maybe one could argue with the absolute continuity of the Lebesgue integral? I find this problem quite interesting though. Please help :-)
Thank's in advance!
Wait isn't this trivial anyway? As the Hausdorff measure can only measure sets which are measureable (and level sets are measureable sets), the inverse image of any set $A\in\mathcal{B}(\mathbb{R})$ has to be measureable as otherwise the image couldn't even exist as the Hausdorff measure wouldn't be able to measure it...