How to prove that the image of a continuous curve in $\mathbb{R}^2$ has measure $0$? How to prove that the image of a continuous curve in $R^2$ has measure $0$?
This is an exercise given in Real Analysis-Stein & Shakarchi. 
A hint is given as follow: Cover the curve by rectangles, using the uniformly continuity of $f$.
 A: Basically, just do what the hint says. The idea is to cover the curve (i.e. the graph of your function $f:\mathbb R \to\mathbb R$) by rectangles such that the sum of their areas (measures) is smaller than $\epsilon>0$ and do so for every $\epsilon>0$. This would show that $m(C)<\epsilon$ for every $\epsilon>0$ and thus $m(C)=0$.
Let's do this.
Lemma. Let $g:[a,b]\to\mathbb R$ be a continuous function and let $\Gamma_g = \{(x,g(x))|\;x\in[a,b]\}$ be the graph of $g$. Then $m(\Gamma_g)=0$.
Proof. The function $g$ is uniformly continuous, since $[a,b]$ is compact. Let $\epsilon>0$. Then there exists a $\delta>0$, such that for each pair of points $x,y\in [a,b]$ such that $|x-y|<\delta$, the inequality $|g(x)-g(y)|<\epsilon$ holds. We may assume without loss of generality that $\delta<b-a$ (since if some $\delta\geq b-a$ works for this choice of $\epsilon$, any other number $0<\delta_1<b-a$ would also work.)
Let $n\in\mathbb N$ be the smallest natural number such that $n\delta> b-a$. Then there exist numbers $a=x_0<x_1<\ldots<x_n=b$ on the interval $[a,b]$, such that for each $i=1,2,\ldots,n$ we have $|x_i-x_{i-1}|<\delta$. (Proving this is an easy exercise.) For each $i=1,2,\ldots,n$ let $y_i=g(x_i)$.
Now, for each $i=1,2,\ldots,n$ and each point $z\in[x_{i-1},x_i]$, we have $g(z)\in[y_i-\epsilon,y_i+\epsilon]$ because for such $z$ the inequality $|z-x_i|<\delta$ holds, and thus by definition of $\delta$, we also have the inequality $|g(z)-y_i|<\epsilon$. This implies that $\Gamma_g\subseteq\bigcup_{i=1}^n[x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon]$ and therefore $$m(\Gamma_g)\leq\sum_{i=1}^n m([x_{i-1},x_i]\times[y_i-\epsilon,y_i+\epsilon])\leq\sum_{i=1}^n 2\delta\epsilon = 2\delta\epsilon n$$ Since $n$ is the smallest natural number for which $n\delta>b-a$, we must also have $b-a+\delta\geq n\delta$ (since otherwise we would have $b-a+\delta<n\delta$ which would imply $b-a<(n-1)\delta$, which would contradict our choice of $n$). This means that $2\delta\epsilon n\leq 2\epsilon(b-a+\delta)<4\epsilon(b-a)$ since $\delta<b-a$.
So, we showed that for every $\epsilon>0$ the inequality $m(\Gamma_g)<4\epsilon (b-a)$ holds. This is only possible if $m(\Gamma_g)=0$ and thus the proof is complete. $\square$
Now, to show this for $f:\mathbb R\to\mathbb R$ notice that the graph of $f$ (defined by $\Gamma_f=\{(x,f(x))|\;x\in\mathbb R\}$) is the union of graphs of restrictions to $[-n,n]$. More precisely: define for each $n\in\mathbb N$ a function $f_n:[-n,n]\to\mathbb R$ by the formula $f_n(x) = f(x)$. Then $$\Gamma_f=\bigcup_{n=1}^\infty\Gamma_{f_n}.$$ Therefore, $$m(\Gamma_f)\leq\sum_{n=1}^\infty m(\Gamma_{f_n})=\sum_{n=1}^\infty 0 = 0$$ and thus $m(\Gamma_f)=0$. We are done.
