I'm struggling with this exercise in Prof Tao's book "Introduction to measure theory".
Exercise 1.2.25 (p.43): Define a continuously differentiable curve in $R^d$ to be a set of the form ${\gamma(t):a \leq t \leq b }$, where $[a,b]$ is a closed interval and $\gamma:[a,b] \mapsto R^d$ is a continuously differentiable function.
i) If $d \geq 2$, show that every continuously differentiable curve has Lebesgue measure zero. (Why is the condition $d \geq 2$ necessary?)
ii) Conclude that if $d\geq2$, then the unit cube $[0,1]^d$ cannnot be covered by countably many continuously differentiable curves.
Can anyone give me some hints?
Since $f$ is continuously differentiable function it is Lipschitz: there exists $M > 0$ with the property that $|f(x) - f(y)| \le M|x-y|$ for all $x,y \in [a,b]$. In particular, if $|x-y| \le \delta$ then $f(y) \in B(f(x),M\delta)$.
Partition $[a,b]$ into $n$ equal subintervals with partition points $t_0,t_1,\ldots,t_n$ and let $m_k$ be the midpoint of $[t_{k-1},t_k]$. Let $\delta = \frac{b-a}{2n}$. Then $f([t_{k-1},t_k]) \subset B(f(m_k),M\delta)$ for each $k$. This leads to $$f([a,b]) \subset \bigcup_{k=1}^n B(f(m_k),M\delta)$$ and in turn the subadditivity of the outer measure implies $$|f([a,b])|_e \le \sum_{k=1}^n C (M\delta)^d = C \left( \frac{b-a}{2} \right)^d n^{1-d}.$$
This holds for all $n \in \mathbb N$.