Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function on the closed interval $[a,b]$. Let $\mathcal{C}=\{(x,f(x))\in\mathbb{R}^2|x\in[a,b]\}$ be its graph. Prove that $\mathcal{C}$ is measurable and $m(\mathcal{C}))=0$.
I have no idea how to start, I don't know how to relate the measurable function and its graph. Can you give me some hints?
Thanks!
The graph is a closed set; hence it is a Borel set.
By Fubini's Theorem $m(\mathcal C)=\int m(\mathcal C_x) dx$ where $\mathcal C_x$ is the section of $\mathcal C$ by $x$. Since $\mathcal C_x$ is the singleton set $\{f(x)\}$ it folows that $m(\mathcal C_x) =0$ for evey $x$. Hence $m(\mathcal C)=0$.
(I am writing $m$ for both the two-dimensional and the one-dimensional Lebesgue measure).
Proof without Fubini's Theorem:
Let $\epsilon >0$. There exists a positive integer $N$ such that $|x-y| \leq \frac 1 N$ implies $|f(x)-f(y)| <\epsilon$. Now you can verify that $\mathcal C \subset \bigcup [x_{i-1},x_i] \times [f(x_i)-\epsilon, f(x_i)+\epsilon]$. It follows that $m(\mathcal C) \leq \sum (x_i-x_{i-1}) \epsilon =\epsilon (b-a)$. Since $\epsilon >0$ is arbitrary we are done.