Hausdorff measure of rectifiable curve equal to its length 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.
 A: To prove $H^1(\bar\gamma)\le L$, begin by picking a partition $t_0,\dots, t_m$ such that 
$$
\sum\limits_{i=1}^md(\gamma(t_{i-1}),\gamma(t_i)) > L-\epsilon
\tag{1}$$
and $d(\gamma(t_{i-1}),\gamma(t_i))<\epsilon$ for each $i$. Let $A_i = \gamma([t_{i-1},t_i])$.
Suppose  $\operatorname{diam} A_i>2\epsilon$ for some $i$. Then there are $t',t''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>2\epsilon$. So, after these numbers are inserted into  the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i)) $ increases by more than $\epsilon$, contradicting $(1)$. Conclusion: $\operatorname{diam} A_i\le 2\epsilon$ for all $i$. 
Suppose $\sum_i\operatorname{diam} A_i>L+ \epsilon$. For each $i$ there are $t_i',t_i''\in (t_{i-1},t_i)$ such that $d(\gamma(t'),\gamma(t''))>\operatorname{diam} A_i - \epsilon/m$.  So, after all these numbers are inserted into  the partition, the sum of differences $d(\gamma(t_{i-1}),\gamma(t_i)) $ will be strictly greater than  $L+\epsilon - \epsilon = L$, which is again a contradiction.
Thus, the sets $A_i$ provide a cover such that $\operatorname{diam} A_i\le 2\epsilon$ for all $i$ and $\sum_i\operatorname{diam} A_i\le L+ \epsilon$. Since $\epsilon$ was arbitrarily small, $H^1(\bar\gamma)\le L$.

For completeness: the opposite direction follows from the inequality $$H^1(E)\ge \operatorname{diam} E\tag{2}$$ which holds for any connected set $E$. To prove it, fix a point $a\in E$ and observe that the image of $E$ under the $1$-Lipschitz map $x\mapsto d(x,a)$ is an interval of  length close to $\operatorname{diam} E$ provided that $a$ was suitably chosen. 
Then apply $(2)$ to each $\gamma([t_{i-1},t_i])$ separately.
A: This statement is proved in "A Course in Metric Geometry" by Dmitri Burago,
Yuri Burago, and Sergei Ivanov:

A: Other approach (it is not completely clear to me that, as the other answer, we can choose such $\epsilon$ in this way) that construct by recursion partitions of $[0,1]$ is as follows
$$
a_{k+1}:=\inf\{x\in[a_k,1]:|\gamma(a_k)-\gamma(x)|=\epsilon\}\cup\{1\}\tag1
$$
where we set $a_0:=0$. Then we have a partition of $[0,1]$ defined by $\mathfrak Z:=\{a_0,a_1,\ldots,a_m\}$ with the property that
$$
\begin{align*}|\gamma(a_k)-\gamma(a_{k+1})|&=\operatorname{diam}\big(\gamma([a_k,a_{k+1}])\big),\quad\forall k\in\{0,\ldots,m-2\}\\
|\gamma(a_{m-1})-\gamma(a_m)|&\le\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\le\epsilon\end{align*}\tag2
$$
(note that by construction $a_m=1$). Then we find that
$$
\mathcal H_\epsilon^1(\bar\gamma)\le\sum_{k=0}^{m-2}\operatorname{diam}\big(\gamma([a_k,a_{k+1}]\big)+\operatorname{diam}\big(\gamma([a_{m-1},a_m])\big)\\
\le\sum_{k=0}^{m-2}|\gamma(a_k)-\gamma(a_{k+1})|+\epsilon\le L(\bar\gamma)+\epsilon\tag3
$$
Then taking limits above as $\epsilon\to 0^+$ we find that $\mathcal H^1(\bar\gamma)\le L(\bar\gamma)$, as desired.
