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.
Best Answer
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.