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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?

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  • $\begingroup$ By "null set" do you mean a set with measure zero? $\endgroup$ – Gregory Grant Dec 8 '15 at 22:22
  • $\begingroup$ Yes, that is what I meant. Should I rephrase? It was originally in a different language. $\endgroup$ – Meitar Dec 8 '15 at 22:24
  • $\begingroup$ Well you don't have to rephrase because it's clarified here in the comments. $\endgroup$ – Gregory Grant Dec 8 '15 at 22:29
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    $\begingroup$ 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). $\endgroup$ – Gregory Grant Dec 8 '15 at 22:30
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Hint.

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 $L. \epsilon$.

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  • $\begingroup$ 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? $\endgroup$ – Meitar Dec 15 '15 at 13:02
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    $\begingroup$ 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. $\endgroup$ – mathcounterexamples.net Dec 15 '15 at 13:21
  • $\begingroup$ 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. $\endgroup$ – Meitar Dec 15 '15 at 13:26
  • $\begingroup$ 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 $\gamma(t)$, $g(t)=\gamma(t)_y$? $\endgroup$ – Meitar Dec 15 '15 at 13:35

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