Csdrhrt

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.

Let $gamma: [0,1]toBbb{R}^n$ for $ngeq 2$ be continuously differentiable. Since it is continuously differentiable, it is Lipschitz continuous with Lipschitz constant $C$, i.e. there exists a $Cgeq 0$ such that for all $t_1,t_2in[0,1]$, $$| x_2-x_1| leq C|t_2-t_1|$$ 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 $i$, $$gamma([t_i,t_{i+1}])subset B(x_i,C|t_{i+1}-t_i|) = Bleft(x_i,frac{C}{k}right) := B_i$$ so that the curve $gamma([0,1]) subset bigcuplimits_{i}B_i$.

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 $$kcdot frac{1}{k^n} = frac{1}{k^{n-1}}$$ Then, as $ktoinfty$, the upper bound for the volume goes to $0$ as long as $ngeq 2$.

Thus, we can create closed ball cover of the curve with arbitrarily small volume, so the curve must be a null set.

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.
$$ [y_j, y_{j+1}] := { tcdot y_{j} + (1-t) y_{j+1} : tin [0,1] }$$
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
$$A_n :={ xin mathbb{R}^2 : dist(x, p_n)leq varepsilon }supseteq Im(gamma)$$
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
$$ L geq sum_{j=0}^{n-1} Vert y_{j+1}-y_j Vert. $$

Think of applying fubini theorem to the characteristic function over the path !