Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^1$ function. I want to prove that $\Gamma_f=\{(x,y)\in \mathbb{R}^{n+1}| y=f(x)\}$ is of measure $0$.
I think that I need to prove that for every rectangle $A$ in $\mathbb{R}^n$, $f|_A$ is a Lipschitz function, but I don't have any ideas about how to do that.
Moreover, I have seen the proof for a continuous function here, but can I say that since every $C^1$ function is definitely $C^0$, so the proof is trivial using the link above?
$m(\Gamma_f)=\int I_{\Gamma_f} dm_{n+1}=\int I_{\Gamma_f}(x,y)dydm_n=0$ since the inside integral is $0$ for each $x \in \mathbb R^{n}$. [I have used $x$ for a vector in $\mathbb R^{n}$ and $y$ for a real variable. I have used Fubini's Theorem to evaluate the integral over $\mathbb R^{n+1}$. I have used the notation $m_n$ for Lebesgue measure on $\mathbb R^{n}$. Also $I_A$ is the characteristic function of $A$: $I_A(z)=1$ if $z \in A$ and $0$ otherwise].
In conclusion we don't need continuity etc. The result is true for any measurable function $f$.