Suppose $\Gamma$ is a curve $y=f(x)$ in $\mathbb{R}^2$, where $f$ is continuous. Show that $m(\Gamma)=0$.
Hint: Cover $\Gamma$ by rectangles, using the uniform continuity of $f$.
If the curve is in $\mathbb{R}^2$, then $y=f(x)$ must be defined from $\mathbb{R}$ to $\mathbb{R}$. Now, if $y=f(x)$ were, hypothetically, mapped from $[a,b]$ to $\mathbb{R}$, where $[a,b] \subset \mathbb{R}$, then I can conclude the fact that $f$ is uniformly continuous since $f$ is continuous over a closed interval. But how do I extend this fact to the domain of $\mathbb{R}$? Because, according to the standard Euclidian topology, $\mathbb{R}$ is viewed as both open or closed.
Write $$\Gamma = \bigcup_{n \in \mathbb{Z}} \Gamma_n$$ where $\Gamma_n$ is the curve $\{(x,f(x)); x \in [n,n+1]\}$. For each fixed $n$, $\Gamma_n$ has measure zero and therefore the countable union $\Gamma$ has measure zero.