# Show that the image of Lipschitz function $\gamma : [0,1] \to R^n$ has measure $0$, if $n \ge 2$.

Problem Statement: Let $\Gamma$ be the image of a Lipschitz continuous function $\gamma : [0,1] \to R^n$, that is, $\Gamma = \{\gamma(t) : t \in [0,1]\}$, and $|\gamma(t_1) - \gamma(t_2)| \le K |t_1 - t_2|$. Show that $m_n(\Gamma) = 0$, where $m_n$ is $n$ dimensional Lebesgue measure.

My attempt at a solution: To start out with, I wanted to check and see if my intuition of why this statement is true makes sense. I believe that the reason this is true is because $m_n([0,1]) = 0,$ since $[0,1] = [0,1] \times \{0\}$ in $R^n$, but I am not sure if that is correct.

Going on, what I have done so far is: proved that since $\gamma$ is Lipschitz, it is absolutely continuous, thus of bounded variation, and can be written as a difference of two increasing functions. Thus, WLOG, assume $\gamma$ is increasing. Now, this implies that if $\{(a_i,b_i)\}_{i=1}^m$ is a finite collection of open intervals that covers $[0,1]$, then $\{f(a_i),f(b_i)\}^m_{i=1}$ covers $\Gamma$.

Now, we know that we can find some such collection of open intervals covering $[0,1]$, such that $m([0,1]) + \epsilon > \sum^m_{i=1}|b_i-a_i|$, and we know that

$$m_n(\Gamma) \le \sum^m_{i=1}|f(b_i)-f(a_i)| \le K \sum^m_{i=1}|b_i-a_i| < 1 + \epsilon.$$

This obviously doesn't get me what I wanted, but I'm not sure how to incorporate the fact that $m_n([0,1]) = 0$... a hint in the right direction would be appreciated.

Hint: For any $m \in \mathbb N,$ $\Gamma$ is contained in
$$\bigcup_{k=1}^m \ B(f(k/m), K/m),$$
where $B(a,r)$ is the closed ball centered at $a$ of radius $r.$
• @mathcounterexamples.net ty for changing $n$ to $m$ in the union.