# Continuous curves and Lebesgue measure

Let $$d\geq 2$$, and let $$\gamma: [a,b]\to \mathbb{R}^d$$ be a continuously differentiable function. Show that the curve of this function $$\{(t,\gamma(t)): a \leq t \leq b\}$$ has Lebesgue measure zero.

This problem was directly or indirectly discussed here, here and here. Nevertheless I have a small confusion regarding my own proof.

I have proven the assertion as follows: by continuiuty assumption we have $$\forall \epsilon>0 \quad \exists \delta>0 \quad \forall x,y \in [a,b]: \quad d(x,y)\leq \delta \implies d(f(x),f(y)) \leq \epsilon$$ Choosing $$\delta$$ for our $$\sqrt[d]{\epsilon}$$ we partition the interval into $$[a, a+\delta],\ldots, [a+(n-1)\delta, a + n\delta]$$ where $$n=\lceil (b-a)/\delta \rceil$$. We then can contain our curve in the boxes (I abbuse the notation a little bit) $$[a, a+\delta]\times[\gamma(a)- \sqrt[d]{\epsilon}\cdot 1, \gamma(a)+ \sqrt[d]{\epsilon}\cdot 1],\ldots, [a+(n-1)\delta, a + n\delta] \times[\gamma(a+(n-1)\delta)- \sqrt[d]{\epsilon}\cdot 1, \gamma(a+(n-1)\delta)+ \sqrt[d]{\epsilon}\cdot 1]$$ each has elementary measure $$n\cdot \delta\cdot\epsilon \approx (b-a) \epsilon$$ and it can get arbitrarily small.

The problem is that I only used continuity and not the continuous differentiability. This is most certainly troubling since in the remark Prof. Tao explicitly says that this result does not hold for continuous functions due to the existence of space filling curves. So at which step in the proof did I make a logical fallacy?

• What you approximated is the measure of the graph of the function, not the image of the function. Commented Jun 11, 2020 at 14:03

More seriously, as I understand it, you're trying to show that the image of $$\gamma$$ has measure zero right? This would mean $$\text{image}(\gamma) \subset \Bbb{R}^d$$, and hence you should be covering it by cubes/rectangles in $$\Bbb{R}^d$$, but you've been covering it using rectangles in $$\Bbb{R}^{d+1}$$, which doesn't make sense. As a result, if you trace through your work, the total measure of the cubes in $$\Bbb{R}^d$$ is $$n \cdot \epsilon$$, not $$n\cdot \delta\cdot\epsilon$$. I'm sure you can see the problem here: as $$\epsilon$$ gets smaller, the $$n$$ could grow much much larger, so the product $$n\cdot \epsilon$$ isn't small anymore. This is the flaw in your proof (another reason why you should suspect your proof is that you never made use of the hypothesis $$d \geq 2$$.).
The typical argument using continuous differentiability allows you to deduce that the function is locally Lipschitz (by the mean-value inequality), and also in this special case, your domain $$[a,b]$$ happens to be compact, so actually the function $$\gamma$$ is Lipschitz (not just locally). This Lipschitz condition gives you a more refined estimate on "how much" your function can grow, and this is exactly what you need to prove the claim.
• I had to show that $\{(t, \gamma(t))\}$ has measure zero, this is why it is in $\mathbb{R}^{d+1}$ Commented Jun 11, 2020 at 16:55
• @CebişMellim are you sure that's what you're asked to prove? Because for that you don't need continuous differentiability; all you need is for the function to be measurable (continuous would work, and I believe your given proof works) (and there's no need for any restrictions on $d$). To prove $\{\gamma(t)| \, \, a \leq t \leq b\}$ has $d$-dimensional Lebesgue measure $0$, we need locally Lipschitz (which is guaranteed by continuous differentiability) and $d\geq 2$. Commented Jun 11, 2020 at 17:03