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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?

Zeno cosini
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  • Instead of Lipschitz, perhaps you can use uniform continuity over compact sets. All you need is continuity for that. – Michael May 30 '19 at 06:31
  • Okay the link you provided asks the same question (but requiring only continuity) and the answers there are as in my comment. So that link already answers your question. Every differentiable function is also continuous. – Michael May 30 '19 at 06:35
  • @Michael Yup! You're right. But I also want to do that using Lipschitz. Don't you have any ideas? Actually, I am solving this for my TA class, and I want to use continouty of it differentially! – Zeno cosini May 30 '19 at 06:45
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    This is an immediate corollary of Fubini's Theorem. Even continuity is not needed. Measurability is enough. – Kavi Rama Murthy May 30 '19 at 07:38
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    @Yasinowski : Take a hypercube of volume 1 in $\mathbb{R}^n$, say the function $f(x)$ has Lipschitz parameter $L$ on that cube. Fix $\epsilon>0$. Chop the cube into disjoint subcubes where points in each subcube are at most $\epsilon/L$ apart. So the variation of $f$ over each subcube is at most $\epsilon$ and the graph in that subcube is in an $(n+1)$-dimensional box of volume $(\mbox{subcube volume})\epsilon$. Summing over all subcubes gives area $\epsilon$. This holds for all $\epsilon>0$ so the graph has measure 0 in the unit cube. $\mathbb{R}^n$ is the union of such unit cubes. – Michael May 30 '19 at 16:36
  • The same argument can be repeated without Lipschitz by just using uniform continuity. – Michael May 30 '19 at 16:42

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$m(\Gamma_f)=\int I_{\Gamma_f} dm_{n+1}=\int I_{\Gamma_f}(x,y)dydm_n=0$ 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 $f$.

Kavi Rama Murthy
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