Exercise 1.2.25 (Lebesgue measure of continuously differentiable curves)

Question

Let $d\geq 2$, and let $\gamma: [a,b]	o \mathbb{R}^d$ be a continuously differentiable function. Show that the curve of this function has Lebesgue measure zero.

Elu Thingol · Accepted Answer

Since $\gamma$ is a continuously differentiable function it must be Lipschitz. In other words, there exists a $M > 0$ with the property that $d(\gamma(x),\gamma(y)) \leq M\cdot d(x,y)$ for all $x,y \in [a,b]$. In particular, if $d(x,y) \leq \delta$ then $f(y) \in B(\gamma(x),M\delta)$.\par

Pick an arbitrary $n\in\mathbb{N}$. 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  for each $k$$$f([t_{k-1},t_k]) \subseteq B(f(m_k),M\delta)$$ 
ewlineThis leads to$$f([a,b]) \subseteq \bigcup_{k=1}^n B(f(m_k),M\delta)$$and in turn the subadditivity of the outer measure implies$$m^*(\gamma([a,b])) \leq \sum_{k=1}^n C (M\delta)^d = C \left( \frac{b-a}{2} ight)^d n^{1-d}$$This holds for all $n \in \mathbb N$, therefore the above quantity must be zero. Taking Cartesian product preserves this by \href{../exercise_1-2-22}{Exercise 1.2.22}.

Exercise 1.2.25 (Lebesgue measure of continuously differentiable curves)

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Exercise 1.2.25 (Lebesgue measure of continuously differentiable curves)

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