TITLE: Hausdorff measure of rectifiable curve equal to its length
QUESTION [5 upvotes]: Let $(\mathbb{R}^n,d)$ be  a metric space. A continuous, injective mapping $\gamma: [0,1]\to \mathbb{R}^n$ is a curve and denote its image $\overline{\gamma}:=\gamma([0,1])$.  I wish to prove that its Hausdorff measure, $H^1(\overline{\gamma})$, is equal to the length of the curve $L$. 
In particular I am having trouble showing that
$$H^1(\overline{\gamma})\leq L.$$
Any ideas?

The length of the curve is defined by
$$L = \sup\left\{\sum\limits_{i=1}^md(\gamma(t_{i-1}),\gamma(t_i))\,\bigg|\, 0 = t_0 < t_1 < \dots < t_m = 1     \right\}. $$
We have that
$$H^1_\delta(E) = \inf\left\{\sum\limits_{i=1}^\infty\text{diam}(A_i)\,\bigg|\,\bigcup\limits_{i=1}^\infty A_i \supseteq E,\,\text{diam}(A_i)\leq \delta\right\}. $$
That is, the infimum is taken over all possible countable coverings $(A_i)_{i=1}^\infty$ of $E$, where the sets $A_i$ are "small enough." We then define the Hausdorff measure as
$$H^1(E) = \lim\limits_{\delta\to 0^+}H^1_\delta(E).$$

My idea is that I want to show that for all $\varepsilon>0$ there exists $\delta > 0$ such that
$$H_\delta^1(\overline{\gamma})\leq L +\varepsilon$$
where $\delta$ is proportional to $\varepsilon$ such that letting $\varepsilon\to 0^+$ also forces $\delta \to 0^+$, and we get
$$H^1(\overline{\gamma})\leq L,$$
however I couldn't succeed in showing this.

REPLY [2 votes]: This statement is proved in "A Course in Metric Geometry" by Dmitri Burago,
Yuri Burago, and Sergei Ivanov: