Cantor curve is not absolutely continuous

Definition : A curve $\omega : [0,1]\to X$ is defined absolutely continuous whenever there exists $g\in L^1([0,1])$ such that $d(\omega(t_0),\omega(t_1))\le\int_{t_0}^{t_1}g(s)ds$ for every $t_0<t_1$.

I would like to show that, according to the above definition, the graph of the Cantor function is not absolutely continuous. I think this is true since I read that for curves $\omega:\mathbb{R}\to\mathbb{R}^d$ we must have that the map $g$ from the definition satisfies $g(t)=\|\dot{\omega}(t)\|$. However, if $\omega$ is the curve of the cantor function, we have $d(\omega(0),\omega(1))=d((0,0),(1,0))=\sqrt{2}$ and $\int_0^1 \|\dot{\omega}(t)\|dt=\int_0^1 \sqrt{1^2+0^2}dt=1$ so $d(\omega(0),\omega(1))\not\leq\int_0^1 \|\dot{\omega}(t)\|dt$.

I was wondering if there is a way to show that the Cantor curve is not absolutely continuous (according to the above definition) in a more direct way without using the result that $g=\|\dot{\omega}\|$? I was thinking by contradiction : "Suppose there exists $g$ with this property..." but I am not sure how this is done.

Maybe it helps to know that the length of the Cantor curve is $2$ from this discussion, so in particular the length of the curve is different from the integral of the velocity.

Edit

Also I was wondering if the definition of absolutely continuous curve whas equivalent to the more natural (for me) condition that the metric derivative exists almost everywhere and for every $a,b$ such that $0\le a \le b\le 1$ we have $$\operatorname{length}_{a,b}(\omega)=\int_a^b|\dot{\omega}|(t)dt$$ where $$\operatorname{length}(\omega):=\sup\left\{\sum\limits_{k=0}^{n-1}d(\omega(t_k),\omega(t_{k+1})),n\ge 1, a=t_0<t_1<\ldots<t_n=b \right\}.$$ and $|\dot{\omega}|(t)$ is the metric derivative defined by $$|\dot{\omega}|(t):=\lim\limits_{h\to0}\frac{d(\omega(t+h,t))}{h}.$$

NB : the definition of absolute continuity has an inequality because although for absolutely continuous function $f:I\to\mathbb{R}$ we have $f(y)-f(x)=\int_x^y f'(t)dt$ this is not true for function $f:[0,2]\to\mathbb{R}^n$. Take $f(t)=(t,0)$ for $t\in[0,1]$ and $f(t)=(1,t-1)$ for $t\in[1,2]$. $d(f(0),f(2))<d(f(0),f(1))+d(f(1),f(2))=\int_0^2 |f'|(t)dt$

Solution 1:

If you have a curve $\gamma:[0,1]\to (X,d)$ where $(X,d)$ is a complete and separable metric space then if $\gamma$ is absolutely continuous with your definition, there always exists the metric derivative and it is what you expect (the least function which you can choose as $g$ in the inequality of absolute continuity). For this fact if you need a proof I can write it.

Now I think that the proof of the fact that the Cantor function is not AC is immediate once you realise that an AC function sends sets of measure zero into sets of measure zero (Lusin N property): however the Cantor function sends the Cantor set to the whole interval $[0,1]$, hence it cannot be absolutely continuous.

EDIT: What we want to prove now is that given $\gamma:[0,1]\to X$ absolutely continuous (and $(X,d)$ complete and separable metric space), then the metric derivative $|\dot{\gamma}_t|:=\lim_{h\to 0}\frac{d(\gamma_{t+h},\gamma_t)}{h}$ exists a.e., it is in $L^1(0,1)$ and it is the minimal $g$ such that $$d(\gamma_t,\gamma_s)\leq\int_s^t|\dot{\gamma}_r|dr\;\;\forall t,s\in[0,1].$$

Pick a countable dense set $(x_n)_n$ in $X$ and define $h_n(t):=d(\gamma_t,x_n)$. By triangle inequality it is immediate to see that $$|h_n(t)-h_n(s)|\leq\int_s^tg(r)dr,$$ where $g$ is your $L^1$ function. By Lebesgue differentiation you also have $|h_n^\prime(t)|\leq g(t)$ for a.e. $t\in[0,1]$. Call $h:=\sup_nh_n^\prime$ and note that it is in $L^1$ thanks to the previous inequality. As an (easy) exercise try to prove that $d(\gamma_t,\gamma_s):=\sup_n(h_n(t)-h_n(s))$. We now have $$d(\gamma_t,\gamma_s)=\sup_n\int_s^th_n^\prime(r)dr\leq\int_s^th(r)dr,$$ so $h$ can substitute $g$ in the definition of absolute continuity. It remains to show that our $h$ is the metric speed: first apply Lebesgue differentiation to the previous inequality to deduce $$\limsup_{t\to s}\frac{d(\gamma_t,\gamma_s)}{|t-s|}\leq h(s)$$ for a.e. $s\in[0,1]$. Then note that by a previous exercise we get $$d(\gamma_t,\gamma_s)\geq h_n(t)-h_n(s) = \int_s^th^\prime_n(r)dr$$ so that we can infer $$\liminf_{t\to s}\frac{d(\gamma_t,\gamma_s)}{|t-s|}\geq h_n^\prime(s)$$ for a.e. $s\in[0,1]$, which allows to conclude.