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I need help to solve the following problem:

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous function. Prove that the graph $G(f)=\{(x,f(x)):x\in\mathbb{R}^n\}$ has measure zero in $\mathbb{R}^{n+1}$.

I suppose that I have to use that f es uniformly continuous, but I don't know what rectangle which sum of volumes is less than $\varepsilon > 0$ should I take.

user326159
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  • Try Fubini. ${}{}{}$ – copper.hat Mar 26 '16 at 22:49
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    I must prove it without using Fubini. Only the definition of measure zero and the properties of a continuous function. – user326159 Mar 26 '16 at 22:50
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    Since it is continuous, you can enclose the graph in a collection of rectangles with arbitrarily small area. Split the domain into compact rectangles and use uniform continuity. Stitch the answer together. – copper.hat Mar 26 '16 at 22:53
  • First show (or use, if accepted as known) that a continuous function on a compact space is uniformly continuous. Then show this proves the graph of the restriction of $f$ to a compact ball has measure zero. Finally, $\mathbb R^n$ is a countable union of compact balls. –  Mar 26 '16 at 22:54
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    You should state your restrictions in the question, otherwise you will waste the time of answerers! – copper.hat Mar 26 '16 at 22:54

3 Answers3

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It is sufficient to show that the set $G'=\{(x,f(x)) | x \in [0,1)^n \}$ has measure zero.

Let $\epsilon>0$, then since $f$ is uniformly continuous on the compact set $[0,1]^n$, there is some $\delta>0$ such that if $\forall x,x' \in [0,1): \|x-x'\|_\infty < \delta$ (note convenient choice of norm) then $|f(x)-f(x')| < \epsilon$.

Now choose $n$ such that ${ 1\over n} < \delta$ and, with $k = (k_1,...,k_n)$, let $x_k = {k \over n}$ and $R_k = \{x_k\} + [0,{1 \over n})^n$, where each of the $k_i$ range through $0,...,n-1$. Note that $\sum_k m R_k = m [0,1)^n = 1$.

Note that for $x \in R_k$, we have $|f(x)-f(x_k)| < \epsilon$, hence $\{ (x,f(x)) \}_{x \in R_k} \subset R_k \times [f(x_k)-\epsilon, f(x_k) + \epsilon]$ and so $m \{ (x,f(x)) \}_{x \in R_k} \le m R_k \cdot m [f(x_k)-\epsilon, f(x_k) + \epsilon] = 2 \epsilon \, m R_k$.

Hence $m G' \le 2 \epsilon \sum_k \, m R_k = 2 \epsilon$.

Since $\epsilon>0$ was arbitrary, we have $m G' = 0$.

Balint Pato
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copper.hat
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  • could I ask why it suffices to show that the $G'$ you defined has measure 0? – homosapien Oct 22 '20 at 17:34
  • @HossienSahebjame The same analysis will apply to any cube of the form $[a,b)^n$. We can cover $\mathbb{R}^n$ with a countable number of cubes of this form and the measure of the union of a countable collection of measure zero sets will itself have measure zero. – copper.hat Oct 22 '20 at 17:44
  • ahh got it thanks! like a countable way of covering the reals? – homosapien Oct 22 '20 at 17:47
  • @copper.hat Sorry for ma follow-up, but may I ask, does $m$ denote measure and do you/we claim every point in $[0,1)^n$ can be written in form of $$\frac{k}n+v, k\in{0,\ldots,n-1}^n,v\in \left[0,\frac1n\right)^n$$? – PinkyWay Mar 14 '22 at 06:48
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    @Invisible Yes & yes. ${k \over n}$ is the 'bottom left' corner of the boxes. – copper.hat Mar 14 '22 at 07:00
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Here's another argument. Assuming the graph is measurable, use Fubini-Tonelli to show that its measure is equal to an iterated integral:

$$ m(G) = \int_{{\mathbb R}^n} \int_{{\mathbb R}} {\bf 1}_{\{f(x)\}}(y) dy dx = \int_{{\mathbb R}^n} 0 dx =0,$$

where the second equality is due to the fact that the Lebesgue measure of the singleton $\{y:y=f(x)\}$ is zero for any $x$.

Now for the measurability of $G$. It's a closed set. Why ? Take $(x,y)$ not in $G$. Then $f(x)\ne y$. Therefore by continuity of $f$, there exists a neighborhood $I$ of $x$ and a neighborhood of $J$ of $y$ such that for all $x \in I$, $f(x)\not\in J$. That is, $I\times J\subset G^c$.

Fnacool
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I thought of a more simpler solution I think. So assume $\epsilon > 0$ to be given. Choose $0<\epsilon_0<\dfrac{\epsilon}{b-a}$. Now, by uniform continuity, $$ \forall \epsilon_0 > 0, \exists \delta(\epsilon_0): \forall x,y \in [a,b], |x-y| < \delta(\epsilon_0) \rightarrow |f(x)-f(y)| < \epsilon_0 $$ so for our chosen $\epsilon_0$, we get a corresponding $\delta(\epsilon_0)$ that satisfies the above.

Now, we want to stack countable number of rectangles on our graph so that they cover the graph and have a total volume less than $\epsilon$. To do this, we would need $n=\dfrac{b-a}{\delta(\epsilon_0)}$ number of rectangles of width $\delta(\epsilon_0)$ arranged edge to edge. Now, we choose the height of each rectangle to be $\epsilon_0$.

Then, all these would cover the entire graph and would have a total volume $$=n \cdot \epsilon_0 \cdot \delta(\epsilon_0)= \dfrac{b-a}{\delta(\epsilon_0)} \cdot \delta(\epsilon_0) \cdot \epsilon_0 = (b-a)\cdot \epsilon_0 < \epsilon$$

Where the last inequality works by our choice of $\epsilon_0$. This satifies the definiton of measure zero. Hence the graph has measure zero.

Meet Chaudhari
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