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.


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.

  • $\begingroup$ Thanks a lot for the answer. I actually found another way to prove the first part, however, I'm interested in your proof of the second inequality. Applying the inequality you wrote I get $$H^1(\gamma([t_{i-1},t_i]))\geq \text{diam}(\gamma([t_{i-1},t_i]))\geq d(\gamma(t_{i-1}),\gamma(t_i)) $$ Summing these up I seem to get something that resembles $L$, but it seems to me that we get inequalities pointing in the incorrect direction. Can you expand a bit on it? $\endgroup$ – Eff Mar 14 '15 at 10:28
  • $\begingroup$ For example, how can one justify that $$H^1(\overline{\gamma}) = \sum\limits_{i=1}^m H^1(\gamma([t_{i-1},t_i])) $$ if it indeed is that which should be used? $\endgroup$ – Eff Mar 14 '15 at 11:57
  • $\begingroup$ $H^1$ is a Borel measure, so it is additive over disjoint Borel sets (and these subarcs are disjoint except for the endpoints, which have measure zero). Summing up, you get $H^1>L-\epsilon$, which is good enough. $\endgroup$ – user147263 Mar 14 '15 at 15:38
  • $\begingroup$ @Meta How do you get a contradiction when proving $\text{diam} A_i \le 2\varepsilon$? $\endgroup$ – Alan Watts Apr 26 '16 at 12:48

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.


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