I got by the following exercise: prove that the image of a continuously differential path $\gamma:[0,1] \rightarrow R^2$ is a null set.
I know that I need to find some cover of intervals $I_j=[a_j,b_j]$ of the image that satisfy $$\sum_{j=1}^\infty |I_j| < \epsilon$$ for any $\epsilon>0$.
However, I am not really sure how to do it.
Any help would be appreciated.
On
Sketch:
Denote by $$ L = \int_0^1 \Vert \gamma'(t) \Vert dt $$ the length of our curve. Let $1>\varepsilon>0$ and pick $\delta>0$ such that $$ \vert x - y \vert < \delta \Rightarrow \Vert \gamma(x) - \gamma(y) \Vert< \varepsilon $$ this we can do as continuous functions on compact sets are uniformly continuous. Now set $y_j=\gamma(\frac{j}{n})$ and denote by $[y_j, y_{j+1}]$ the segment between $y_j$ and $y_{j+1}$, i.e. $$ [y_j, y_{j+1}] := \{ t\cdot y_{j} + (1-t) y_{j+1} \ : \ t\in [0,1] \}$$ Set $$ p_n := \bigcup_{j=0}^{n-1} [y_j, y_{j+1}].$$ This is a piecewise linear interpolation of our curve. Now we thicken this interpolation up, in such a way that it contains the original curve. For this pick $n>\frac{1}{\delta}$, then by uniform continuity every point on the image of the curve has distant at most $\varepsilon$. Therefore, we have $$A_n :=\{ x\in \mathbb{R}^2 \ : \ dist(x, p_n)\leq \varepsilon \}\supseteq Im(\gamma)$$ Now we are left to show that the measure of $A_n$ is small. Indeed, I leave it to you to show that we have $$ \vert A_n \vert \leq \varepsilon \sum_{j=0}^{n-1} \Vert y_{j+1} - y_j \Vert + \varepsilon^2 \cdot \pi \leq (L+\pi ) \varepsilon$$ The idea behind the first inequality is that you thicken up in the orthogonal direction, thus you get the length of the segment times the maximal distance ($=\varepsilon$) plus you get halves of balls at the beginning and the end of the linear interpolation. For the second inequality we used that $\varepsilon<1$ (and thus $\varepsilon^2 <\varepsilon$) and the inequality $$ L \geq \sum_{j=0}^{n-1} \Vert y_{j+1}-y_j \Vert. $$
Let $\gamma: [0,1]\to\Bbb{R}^n$ for $n\geq 2$ be continuously differentiable. Since it is continuously differentiable, it is Lipschitz continuous with Lipschitz constant $C$, i.e. there exists a $C\geq 0$ such that for all $t_1,t_2\in[0,1]$, $$\| x_2-x_1\| \leq C|t_2-t_1|$$ where $x_i = \gamma(t_i)$. This implies that $$\gamma([t_1,t_2])\subset B(x_1,C|t_2-t_1|),$$ a closed ball centered at $x_1$ with radius $C|t_2-t_1|$.
Take $k+1$ equally spaced points along $[0,1]$ with $$0=t_0 < t_1 < \dots < t_{k-1} < t_k = 1$$ Then, for each $i$, $$\gamma([t_i,t_{i+1}])\subset B(x_i,C|t_{i+1}-t_i|) = B\left(x_i,\frac{C}{k}\right) := B_i$$ so that the curve $\gamma([0,1]) \subset \bigcup\limits_{i}B_i$.
The volume of a ball with radius $r$ in $\mathbb{R}^n$ is proportional to $r^n$, so the volume of each $B_i$ is proportional to $\frac{1}{k^n}$. Then, the sum of the volumes is an upper bound for the volume of the union, and it is proportional to $$k\cdot \frac{1}{k^n} = \frac{1}{k^{n-1}}$$ Then, as $k\to\infty$, the upper bound for the volume goes to $0$ as long as $n\geq 2$.
Thus, we can create closed ball cover of the curve with arbitrarily small volume, so the curve must be a null set.