# Hausdorff Measure- Lower semi-continuity

By definition , when we are given a set $A \in \mathbb{R}^n$ , $$H_\delta^{n-1} (\partial A ) = \inf \left\{ \sum_{j=1}^{\infty} \alpha_{n-1}\frac{1}{2^{n-1}} [\operatorname{diam}(U_j)] ^{n-1} \mid \partial A \subseteq \cup U_j , \operatorname{diam}U_j \leq \delta \right\} ,$$ $$H^{n-1} (\partial A ) = \lim_{\delta \to 0 } H_\delta^{n-1} (\partial A ).$$

How can I prove that when I have $\{A_i \},A$ which are open with smooth boundaries in $\mathbb{R}^n$ , such that $\operatorname{Leb}(A_i \Delta A ) \to 0$ , then $H^{n-1} ( \partial A) \leq \liminf_{ i \to \infty} H^{n-1} (\partial A_i )$ .

Thanks in advance !

p.s.- $\alpha_{n-1}$ is the volume of the $n-1$ dimensional unit ball

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Why do you think it's true and what are your thoughts so far? Roughly speaking, my thoughts are this: Consider only a bounded portion of $\partial A$. Note that you can take the sets in the definition of Hausdorff measure to be open. Then I would take a covering $U_j$ of $\partial A$ by open sets and using the fact that $|A_i\Delta A| \to 0$, try to show that for sufficiently large $i$, $U_j\cap A$ is a covering of $A_i$. –  Spencer Jul 11 '12 at 11:25
Actually I think I found it on google when I was trying to find properties of the Hausdorff Measure... I have tried proving it using the way you suggested, but without any success. Have you got any other ideas that might help? –  joshua Jul 11 '12 at 14:18

For sets with smooth boundary the Hausdorff measure of the boundary agrees with their perimeter. The perimeter can be defined in at least three different but equivalent ways, two of which you will find on Wikipedia. But I will use the third, known as Miranda's definition: the perimeter of set $A$ is the infimum of $\liminf \|\nabla g_n\|_{L^1}$ taken over all sequences of smooth functions $g_n$ which converge to $\chi_A$ in the $L^1$ norm.
Your assumption says that the characteristic functions of $A_i$ converge to the characteristic function of $A$ in the $L^1$ norm. Take the appropriate sequences $g_{n,i}\to \chi_{A_i}$, and the diagonal sequence $g_{n,n}$ will give the desired inequality.