[Math] Prove that the image of a curve has zero content

analysiscurvesintegrationmeasure-theorymultivariable-calculus

Definition: A set $A \subset \mathbb{R}^2$ is said to have zero content if, for all given $\varepsilon >0$, exists a finite collection of rectangles $A_1, \dots, A_n$ such that $A \subset \bigcup_{i=1}^n A_i$ and
$$\sum_{i=1}^n m(A_i)< \varepsilon,$$
where $m(A_i)$ is the area of the rectangle $A_i$.

Given a $C^1$ curve $\gamma:[a,b] \to \mathbb{R}^2$, show that the image $\gamma([a,b])$ of $\gamma$ has zero content.

This is an exercise in calculus that my teacher asked us to do. We're studying double integrals and the theorem which states that a continuous limited function which border of its domain has zero content is integrable over this domain. So… I'd like to have a proof in this sense (of calculus). I don't wanna use some "huge" result of measure theory or something like this… I've tried to work with partitions of the interval $[a,b]$ but without any success… How to find the required collection of rectangles $A_i$!? Need some help!

PS: I've been successful in proving the particular case where $\gamma([a,b])$ is a graph of a function $f:[c,d]\to \mathbb{R}$ for some interval $[c,d]$ by using that $f$ is continuous and therefore integrable and working with inequalities based on
$$\left|\sum_{i=1}^nf(c_i)\Delta x_i-\int_c^df(x)dx\right|< \varepsilon$$
where $\sum_{i=1}^nf(c_i)\Delta x_i$ is the Riemann Sum relative to any partition $P:c=x_0<x_1<\dots<x_n=d$ which $\max \Delta x_i$ is less than a certain $\delta>0$ which depends only of $\varepsilon$ but not of the particular choice of $c_i \in [x_{i-1},x_i]$. (Choose then $s_i,t_i \in [x_{i-1},x_i]$ such that $f(s_i)=\max_{x\in[x_{i-1},x_i]}f(x)$ and $f(t_i)=\min_{x\in[x_{i-1},x_i]}f(x)$ and work with triangle inequalities!)

Best Answer

If you already proved the particular case when the curve is a graph, then you can use the implicit function theorem!

By the theorem, every $C^1$ curve is (locally) a graph, either of the form $(x, f(x))$ or of the form $(f(y),y)$.

So, your proof can be expanded as follows:

  1. Take a curve $(x(t),y(t))$, where $t\in[a,b]$.
  2. For every point $t_0\in [a,b]$, there exists some $\delta_{t_0}$ such that the $(x(t),y(t))$, $t\in (t_0-\delta_{t_0}, t_0+\delta_{t_0})$, can be parametrized as $(x,f(x))$ or as $(f(y),y)$.
  3. Since $$[a,b] = \bigcup_{t\in[a,b]} (t-\delta_t, t+\delta_t),$$ and since $[a,b]$ is comact, there exists a finite covering of $[a,b]$, i.e. there exist $t_1,t_2,\dots,t_n$ such that $$[a,b]=\bigcup_{i=1}^n (t_i-\delta_{t_i}, t_i+\delta_{t_i}).$$
  4. On each interval $(t_i-\delta_{t_i}, t_i+\delta_{t_i})$, you can cover the curve with pieces of a total area of less than $\frac{\epsilon}n$.
  5. Therefore, you can cover the whole curve with an area of less than $n\cdot\frac\epsilon n = \epsilon$
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