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The book I'm reading is Mathmatical Analysis II by Zorich. (And I havn't learnt any about Real Analysis). In 11.1.2 it proves that the graph of a continous function $f:I\to \mathbb R$($I\subset\mathbb{R}^{n-1}$ is a (n-1)-dimentional interval) is of n-dimentional measure zero. Then a Remark follows:

one can conclude that in general the graph of a continous function $f:\mathbb R^{n-1}\to R$ or a continous function $f:M\to \mathbb R^{n-1}$, where $M\subset \mathbb{R}^{n-1}$, is a set of n-dimentional measure zero.

Let $f:E\to \mathbb{R}(E\subset\mathbb R^{n-1})$ be a continous function, $G(f):=\{(x,y)|x\in E,y=f(x)\}$ is the graph of $f$. I wanna know if $G(f)$ is a set of n-dimentional measure zero (in the Lebesgue sense) in $\mathbb R^n$?

Here is what I've got so far:

(1)If $E$ is a compact set (e.g.a cube) in $\mathbb R^{n-1}$, then it's graph is of measure zero.

(2)If $E$ is open, then $E$ can be represented as the union of a countable number of cubes (no two of which have any interior points in common). So $G(f)$ is the union of a countable number of sets of measure zero, hence $G(f)$ is of measure zero.

(3)If $E$ is closed, the set $E\cap I_i$ is compact, where $I_i\subset \mathbb R^{n-1}$ is a cube and $\bigcup_{i\in\mathbb N} I_i=\mathbb R^{n-1}$. Hence $G(f)=\bigcup_{i\in\mathbb N}E\cap I_i$ is of measure zero.

(4)If $E$ is addmissible, then $E\subset\partial E\cup \mathring{E}$, since $\mathring E$ is open and $\partial E$ is of measure zero, $E$ is also of measure zero.

What if $E$ is any set? For example, for $E=\mathbb R^{n-1}\setminus\mathbb Q^{n-1}$, is $G(f)$ still of measure zero? If not, could you give a counterexample? And what properties should $E$ satisfy to let $G(f)$ be of measure zero?

William Leynoid
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    If you are familiar with Fubini-Tonelli's theorem, then you can apply it to get the desired result. $\int_{R\times R^{n-1}}\mathbb{1}{{(x,f(x)}}(u,v),du,dv=\int_R\Big(\int{R^{n-1}}\mathbb{1}_{{f(u)}}(v),dv\Big),du=0$ since the Lebesgue measure (in $R^{n-1}$) of a single point ${f(u)}$ is zero. – Mittens Jul 20 '21 at 15:07
  • I havn't learnt lebesgue integral, and I think this question has an elementary proof which only relates to the definition of measure zero. – William Leynoid Jul 20 '21 at 15:14
  • In such case, it is enough to concentrate on that graph of $f$ with $f$ restricted to $n-1$-dimensional box and show that each such piece of the graph has measure zero (the whole graph can then be covered be a countable union of pieces of measure zero). – Mittens Jul 20 '21 at 15:19
  • But the question is how to show each piece of graph has measure zero when $E$ is any set? – William Leynoid Jul 20 '21 at 15:28
  • All the same, just concentrate on countable pieces of $E$ (say the intersection of $E $ and a box$. tat way you will be able to control the rise of the little cubes you require to cover a piece of the graph. – Mittens Jul 20 '21 at 15:32
  • $E\cap I_i $ is not necessarily compact, so the restriction of $f$ is not always uniformly continuous, so. how to control the rise exactly? – William Leynoid Jul 20 '21 at 15:40
  • If you do not know the Lebesgue integral, what is your definition of an $n$-dimensional measure? Note that there are several inequivalent definitions, for instance, inner measure, outer measure. I did not read Zorich's book for many years and have no recollection what notion of a measure he is using. From what you wrote, you are likely not ready to solve this exercise. – Moishe Kohan Jul 20 '21 at 16:09
  • The book only mentioned the definition of measure zero. A set E is of measure zero means $\forall \epsilon>0,\exists I_i, i\in\mathbb N,E\subset\bigcup I_i, \sum|I_i|<\epsilon$. And this question is only about measure zero. – William Leynoid Jul 20 '21 at 16:23

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