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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 $A_i$.2012-07-11
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    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?2012-07-11

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