I started studying multivariable integration and still trying to grasp the conecpt of the measure. I`m doing excersices and I keep getting the feeling im doing something wrong so I hope one of you could help me :

Definition: $E \subset \mathbb{R}^n$ is neglectible if for every $\epsilon$ >0 there exists a series of open cubes $Q_1,Q_2,...$ such that $E\subset \cup Q_i$ and $\sum_1^\infty V(Q_i)<\epsilon$

Now, I`m trying to show that if $\phi:[0,1]\to \mathbb{R}^2$ a continously differentiable path, then $\phi([0,1])$ is neglectible.

My idea was to say that $\phi$ is continously differentiable on compact set and hence lipchitz, so we have : $d(\phi(x),\phi(y))\leq Ld(x,y)$.

Now, [0,1] is compact so we can cover it by finitely many open sets of diameter $\sqrt{ \epsilon/L}$,

Excplicitly: $[0,1]\subset \cup_1^NI_i$. Now we can define a cover for $\phi([0,1])$ as folllowing :

For each $i$, take $Q_i$ coverage of $\phi(I_i)$. Its sides are bounded from the inequality : $d(\phi(x),\phi(y))\leq Ld(x,y)<\sqrt{ \epsilon/L}$, so overall $V(Q_i)<\epsilon$

To finish off the proof - and this is where my problem is :

$\phi([0,1])\subset \cup_1^N\phi(I_i)$ $\implies$ $V(\phi([0,1]))\leq \sum_1^NV(Q_i)=N\epsilon$

However- N is the number of open sets required to cover $[0,1]$, so it obviously depends on $\epsilon$ and most likely to get large as $\epsilon$ gets smaller.

I can't understand what I'm missing here. I'll be glad if someone can help me.

  • $\begingroup$ Can you write $N$ in terms of $\epsilon$? $\endgroup$ – Uskebasi Dec 10 '18 at 9:44
  • $\begingroup$ Youre right, its about $1/ \epsilon$ , Which still isn`t good $\endgroup$ – Sar Dec 10 '18 at 9:48
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    $\begingroup$ Isn't it $\frac{\sqrt{L}}{\sqrt{\epsilon}}$? $\endgroup$ – Uskebasi Dec 10 '18 at 9:50
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    $\begingroup$ Aha! True. Thanks alot . $\endgroup$ – Sar Dec 10 '18 at 9:52
  • $\begingroup$ @Uskebasi You should post that as an answer. It is better to not leave questions answered only in the comments. $\endgroup$ – Brahadeesh Dec 10 '18 at 18:41

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