# Image of a continuously differentiable curve $\gamma:[0,1]\to \Bbb{R}^2$ is a null set

Image of a continuously differentiable curve $\gamma:[0,1]\to \Bbb{R}^2$ is a null set. I do know that $\Gamma_{\gamma}=\{(x,\gamma(x)):x\in [0,1]\}$ is null in $R^3$ following a theorem in class. This is the part where I am not quite sure in what I am doing. Let $\epsilon >0$. There exist open bricks $Q_j=I^j_1\times I^j_2\times I^j_3$ such that $\sum v(Q_j)=\sum |I^j_1|\cdot |I^j_2|\cdot |I^j_3|<\epsilon$ and $\Gamma_{\gamma}\subseteq\cup Q_j$. I was about to say that if $Q'_j=I^j_2\times I^j_3$ then $\gamma([0,1])\subseteq \cup Q'_j$ and $\sum v(Q'_j)\le \sum v(Q_j)<\epsilon$ but that is not necessary at all. How can show the set is null?

• By "null set" do you mean a set with measure zero? – Gregory Grant Dec 8 '15 at 22:22
• Yes, that is what I meant. Should I rephrase? It was originally in a different language. – Meitar Dec 8 '15 at 22:24
• Well you don't have to rephrase because it's clarified here in the comments. – Gregory Grant Dec 8 '15 at 22:29
• You definitely have to use the fact that it's differentiable because it's not true otherwise. A continuous function from $[0,1]$ to $\Bbb R$ can be surjective (onto). – Gregory Grant Dec 8 '15 at 22:30

As $\gamma$ is continuously differentiable it has a finite length $L$. Also $\Vert \gamma^\prime \Vert$ is bounded on $[0,1]$, let's say by $M$.
For $\epsilon >0$, pick-up a polygonal line $\gamma_\epsilon$ which approximate $\gamma$ such that $$\sup\limits_{t \in [0,1]} \Vert \gamma(t)-\gamma_\epsilon(t) \Vert \le \epsilon$$ Then the measure of $\gamma([0,1])$ is less than $L. \epsilon$.
• I didn't understand the last expression. Would you explain what you meant by $L.\epsilon$? $L\cdot \epsilon$ or $L-\epsilon$? Or something else? – Meitar Dec 15 '15 at 13:02
• It's the product of $L$ by $\epsilon$. It means that the curve lies inside a tube of length $L$ and a radius of $\epsilon$. Hence the bound for its measure that can be as small as desired. – mathcounterexamples.net Dec 15 '15 at 13:21
• I think it uses tools I haven't still studied. We also didn't yet learn how to integrate and stuff in $\Bbb{R}^n$. Might there be a less advanced way? I do get the main idea and it does make sense to me, but I am having a hard time formulating it as I see I am to use things inconsistent with the recent course notes. – Meitar Dec 15 '15 at 13:26
• What about presenting $\gamma([0,1])$ as a graph of a continuously differentiable curve $g$ from the interval $a,b$ where $a$ is the minimal $x$ coordinate of $\gamma(t)$ over $[0,1]$ and $b$ the same idea such that the image of $t\in [a,b]$ is the $y$ coordinate of $\gamma(t)$, $g(t)=\gamma(t)_y$? – Meitar Dec 15 '15 at 13:35