$\| \gamma(b) - \gamma (a) \| \leq \mathcal{H}^1(\gamma([a,b]))$ if $\gamma$ is not rectifiable?

Question

I'm dealing with Hausdorff-Measure and the length of curves. It widely known that for a rectifiable continuous curve $\gamma:[a,b] \rightarrow \mathbb{R}^n$ the inequasion $$\| \gamma(b) - \gamma (a) \| \leq \mathcal{H}^1(\gamma([a,b]))$$ holds. But does it also hold if $\gamma$ ist not necessarily rectifiable?

Answer 1

Consider the following theorem from here:

Theorem. A set $E\subseteq\mathbb{R}^n$ is the image of a rectifiable curve if and only if it is compact, connected, and $\mathcal{H}^1(E)<\infty$.

As $\gamma:[a,b]\to\mathbb{R}^n$ is continuous, we know that $\gamma([a,b])$ is both compact and connected.

Now consider two cases: $\gamma([a,b])$ has a rectifiable parameterization, or it does not.

In case it does not, then the above theorem tells us that $\mathcal{H}^1(\gamma([a,b]))=\infty$, and so the inequality trivially holds.

If it does, denote this by $\zeta:[a,b]\to\mathbb{R}^n$, and find $a',b'\in[a,b]$ such that $\zeta(a')=\gamma(a)$ and $\zeta(b')=\gamma(b)$. Suppose without loss of generality that $a'<b'$. Then

$$\lVert\gamma(b)-\gamma(a)\rVert=\lVert\zeta(b')-\zeta(a')\rVert\leq\mathcal{H}^1(\zeta([a',b']))\leq\mathcal{H}^1(\zeta([a,b]))=\mathcal{H}^1(\gamma([a,b])),$$

and the inequality holds.