Lower semi-continuity of one dimensional Hausdorff measure under Hausdorff convergence Let $\mathcal H^1$ be the one-dimensional Hausdorff measure on $
\mathbb R^n$, and let $d_H$ be the Hausdorff metric on compact subsets of $\mathbb R^n$. If $K_n$ is connected for all $n \in \mathbb N$, and $d(K_n,K) \to 0$, I would like to know if 
$$\mathcal H^1 (K) \leq \liminf\limits_{n \to \infty} \mathcal H^1(K_n).$$
If $K_n$ is not connected, this is not true. One can take $K_n = \bigcup\limits_{i=0}^{2^n-1} [{i \over 2^n}, {i + 1/2 \over 2^n}]$. Then $\mathcal H^1(K_n) = 1/2$, but $K_n \to [0,1]$. Also, for $\mathcal H^k$, $k$ an integer greater than one, this fails spectacularly. See this picture from Frank Morgan's book: 
The thing that I am trying to prove is used implicitly in Peter Jones's paper on the analyst's traveling salesman problem, I believe.
 A: This is called Golab's Theorem. Two different proofs can be found in


*

*The Geometry of Fractal Sets by K. J. Falconer, Theorem 3.18.

*Topics on Analysis in Metric Spaces by L. Ambrosio and P. Tilli, Theorem 4.4.17


I'll sketch the proof following Falconer. 
Step 1: $K$ is connected. Indeed, if $K$ is partitioned into two nonempty compact sets, then their disjoint neighborhoods form a separation of $K_n$ for $n$ large enough.
Step 2: If $\liminf$ is infinite, there is nothing to prove. Otherwise, we may assume $\mathcal H^1(K_n)\le C<\infty$, by passing to a subsequence. 
Step 3: Replace each $K_n$ with a topological tree $T_n\subset K_n$ such that $T_n$ still converges to $ K$. To this end, pick a finite $1/n$-net in $K_n$ and join its points by arcs, one at a time, without creating loops. This uses the fact that a continuum of finite $\mathcal H^1$ measure is arcwise connected (Lemma 3.12 in Falconer's book).
Step 4: Fix $\delta>0$ and decompose each $T_n$ into the union $\bigcup_{j=1}^k T_{nj}$ of continua  with diameter at most $\delta$ such that $\sum_{j=1}^k \mathcal H(T_{nj}) = \mathcal H^1(T_n)$. This takes another lemma, proved by repeatedly truncating the longest branches of the tree. It's important that $k$ can be chosen independently of $n$; it depends only on $\delta$ and $\sup_n\operatorname{diam}T_n$.
Step 5: Using the Blaschke selection theorem (a sequence of compact sets within a bounded set has a convergent subsequence), and taking some subsequences, we can arrange that $T_{nj}\to E_j$ as $n\to\infty$, for each $j$.
Step 6: Since $K=\bigcup_{j=1}^k E_j$ and $\operatorname{diam}E_j\le \delta$ for each $j$, it follows that 
$$\mathcal H^1_\delta(K)\le \sum_{j=1}^k \operatorname{diam}E_j 
= \lim_{n\to\infty} \sum_{j=1}^k \operatorname{diam}T_{nj} 
\le \liminf_{n\to\infty} \sum_{j=1}^k \mathcal H^1 (T_{nj}) \\
\le \liminf_{n\to\infty} \mathcal H^1 (T_{n}) 
\le \liminf_{n\to\infty} \mathcal H^1 (K_{n})  $$
as claimed. This uses the continuity of diameter with respect to the Hausdorff metric.
