Show that differentiable curves have measure zero in R^2

[Demon117]

Homework Statement

(a) Let $\alpha:I=[a,b]→R^2$ be a differentiable curve. Assume the parametrization is arc length. Show that for $s_{1},s_{2}\in I, |\alpha(s_{1})-\alpha(s_{2})|≤|s_{1}-s_{2}|$ holds.

(b) Use the previous part to show that given $\epsilon >0$ there are finitely many two dimensional open discs $B_{\epsilon}(x_{i}), i=1,..,n$ such that $\alpha (I)\subset \cup _{i=1..n}B_{\epsilon}(x_{i})$ and $\sum_{i}Area(B_{\epsilon}(x_{i}))<\epsilon$.

2. The attempt at a solution

(a) For this I made an argument using the mean value theorem for equality. For all $s\in I$, we have $|\alpha '(s)|=1$ since the curve is parametrized by arc length. Then, given some $s_{o}\in (s_{1},s_{2})$ for some $s_{1},s_{2}\in[a,b]$ we have by the mean value theorem

$\frac{\alpha(s_{1})-\alpha(s_{2})}{s_{1}-s_{2}}=\alpha'(s_{o})$

whereby,

$\frac{|\alpha(s_{1})-\alpha(s_{2})|}{|s_{1}-s_{2}|}=1$, so equality holds.

I am unsure how I should show that it has to be less than. I will have to think of this more. What I am really stuck on is how to even attempt part (b). Any suggestions would be very appreciated.

 

[micromass]

You seem to be applying the mean-value theorem like $\alpha$ was a function in one-variable. But in reality, it is a vector-valued function. Do you know the mean-value theorem for vector-valued functions?

 

[Demon117]

You seem to be applying the mean-value theorem like $\alpha$ was a function in one-variable. But in reality, it is a vector-valued function. Do you know the mean-value theorem for vector-valued functions?

Actually, I hadn't caught that until you pointed it out. I am not familiar with that theorem.

 

[micromass]

Actually, I hadn't caught that until you pointed it out. I am not familiar with that theorem.

OK, doesn't matter, you can do it with integrals easily.

LeTake $v$ a vector with $\|v\|= 1$. Show that

$$\|\alpha(b) - \alpha(a)\| = \int_a^b \alpha^\prime(t)\cdot v ~dt \leq \int_a^b\|\alpha^\prime(t)\|dt$$

then take a suitable $v$ to prove your conjecture.

 

[Demon117]

OK, doesn't matter, you can do it with integrals easily.

LeTake $v$ a vector with $\|v\|= 1$. Show that

$$\|\alpha(b) - \alpha(a)\| = \int_a^b \alpha^\prime(t)\cdot v ~dt \leq \int_a^b\|\alpha^\prime(t)\|dt$$

then take a suitable $v$ to prove your conjecture.

Ok, so I have taken care of that. Thanks for the suggestions, they were quite helpful. I have also shown the second part quite well enough. The only thing that remains is the following:

(c) If h is a bounded continuous function on $R^{2}$, using the previous part, give a reasonable argument as to why

$∫_{\alpha}h dx dy=0$

should hold.

The thought I had was to use the change of variables theorem to integrate over a larger set containing $\alpha(I)$. But I suppose you would have to define some local parametrization in $R^2$, unless you took the union of the disks as in part (b) to be the locally parametrized surface? That seems really odd though and doesn't make a lot of sense. . .

 

[micromass]

So in general, if $E$ is any set of measure $0$ and if $f$ is some function (let's say continuous), then

$$\int_E f = 0$$

Try something like

$$\left|\int_E f\right| \leq \int_E |f|\leq \int_E \textrm{sup}(|f|)\leq \lambda(E)\textrm{sup}(|f|)$$

 

[Demon117]

So in general, if $E$ is any set of measure $0$ and if $f$ is some function (let's say continuous), then

$$\int_E f = 0$$

Try something like

$$\left|\int_E f\right| \leq \int_E |f|\leq \int_E \textrm{sup}(|f|)\leq \lambda(E)\textrm{sup}(|f|)$$

What is the significance of $\lambda(E)$ in this case? Could you define that for me? It seems rather arbitrarily chosen to me.

 

[micromass]

What is the significance of $\lambda(E)$ in this case? Could you define that for me? It seems rather arbitrarily chosen to me.

The function $\lambda$ is the Lebesgue measure. Have you not seen this?

 

[Demon117]

The function $\lambda$ is the Lebesgue measure. Have you not seen this?

Yes, now that makes sense. Our notation uses m rather than $\lambda$. I appreciate your help.