Measure of $C^1$ path in $\mathbb{R}^2$

I started studying multivariable integration and still trying to grasp the conecpt of the measure. Im 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, Im 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.

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