I have a question about some calculation which is in the paper : Two counterexamples in low-dimensional length geometry - Burago, Ivanov, and Shoenthal
I want to say their second result and the calculation is omitted in the paper
I will say roughly their second result : Note that if $c_n :[0,1]\rightarrow (X,d)$ is a continuous curve where $X$ is a length metric space and if $c_n$ goes to $c$ uniformly then we have $${\rm length} \ c\leq {\rm lim\ inf}_n \ {\rm length}\ c_n\ \ast$$
But they have a counter example s.t. $2$-dimensional Hausdorff measure $h_2$ does not satisfy property like $\ast$
my answer :
i) Divide $[0,1]^2$ into 9 squares of equal sides so that we delete five. So we have four squares around vertices of $[0,1]^2$.
If it is denoted as $C_1$, then we do the process repeatedly so that we define $C:=\bigcap_i\ C_i$ (cf. Cantor Set)
ii) In $\mathbb{R}^3$, consider a closed disks $D_i$ s.t. $D_i \rightarrow D$ in Gromov-Hausdorff distance, where $D$ is still a disk.
Assume that $D_i$ has empty intersection with $xy$-plane and $D$ has $C$ intersection with $xy$-plane.
iii) Assume that $2\varepsilon> {\rm area}\ D_i>\varepsilon$ for all $i$.
Since $D$ contains $C$, so ${\rm area}\ D\geq \delta +\varepsilon$ for some $\delta > 2\varepsilon$.