# Is the graph of a continous function $f:E\to\mathbb{R}$($E\subset\mathbb R^{n-1}$) a set of n-dimentional measure zero?

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?

• 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. – Oliver Diaz Jul 20 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. – LEY Jul 20 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). – Oliver Diaz Jul 20 at 15:19
• But the question is how to show each piece of graph has measure zero when $E$ is any set? – LEY Jul 20 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. – Oliver Diaz Jul 20 at 15:32