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Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function on the closed interval $[a,b]$. Let $\mathcal{C}=\{(x,f(x))\in\mathbb{R}^2|x\in[a,b]\}$ be its graph. Prove that $\mathcal{C}$ is measurable and $m(\mathcal{C}))=0$.

I have no idea how to start, I don't know how to relate the measurable function and its graph. Can you give me some hints?

Thanks!

Rookie
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1 Answers1

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The graph is a closed set; hence it is a Borel set.

By Fubini's Theorem $m(\mathcal C)=\int m(\mathcal C_x) dx$ where $\mathcal C_x$ is the section of $\mathcal C$ by $x$. Since $\mathcal C_x$ is the singleton set $\{f(x)\}$ it folows that $m(\mathcal C_x) =0$ for evey $x$. Hence $m(\mathcal C)=0$.

(I am writing $m$ for both the two-dimensional and the one-dimensional Lebesgue measure).

Proof without Fubini's Theorem:

Let $\epsilon >0$. There exists a positive integer $N$ such that $|x-y| \leq \frac 1 N$ implies $|f(x)-f(y)| <\epsilon$. Now you can verify that $\mathcal C \subset \bigcup [x_{i-1},x_i] \times [f(x_i)-\epsilon, f(x_i)+\epsilon]$. It follows that $m(\mathcal C) \leq \sum (x_i-x_{i-1}) \epsilon =\epsilon (b-a)$. Since $\epsilon >0$ is arbitrary we are done.

Kavi Rama Murthy
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  • INTERESTING, so there are continuous function whose graphs has Hasudorff dimension 2 but null Lebesgue measur (https://mathoverflow.net/questions/23960/how-big-can-the-hausdorff-dimension-of-a-function-graph-get) – Tito Eliatron Mar 31 '20 at 07:48
  • I am studying Real Analysis written by Stein, but I didn't reach Lebesgue integration yet. It's a question shown in my assignment. If the solution is related to integration, it goes beyond too far from now what we are studying. Anyway, thanks. – Rookie Mar 31 '20 at 08:06
  • @Tito Eliatron: See also How big can the Hausdorff dimension of a function graph get? Incidentally, most (if not all) of the explicit constructions give continuous functions whose graphs are such that each of its nonempty intersection with an open set in ${\mathbb R}^2$ has Hausdorff dimension $2$ (i.e. the graph is "everywhere locally of Hausdorff dimension $2$"). That is, the graph is "Hausdorff dimension $2$ dense". – Dave L. Renfro Mar 31 '20 at 08:06
  • @YangGao I have added a proof which avoids integration. – Kavi Rama Murthy Mar 31 '20 at 08:17
  • @DaveL.Renfro I konw that (even more, I have a paper on it https://www.tandfonline.com/doi/abs/10.1080/03081087.2019.1612832?journalCode=glma20 https://personal.us.es/bassas/doc/papers/GLMA_preprint.pdf) Simply I don't realize tilll now the relation between Hausdorff dimension an Lebesgue measure. Moreover, if speak about continuous curves There are "many" space filling continuous curves (https://www.sciencedirect.com/science/article/pii/S0024379514007381?via%3Dihub https://arxiv.org/abs/1407.3951) with 2-dim non nul Lebesgue measure. – Tito Eliatron Mar 31 '20 at 08:20
  • I noticed that the link Gjergji Zaimi gave at the mathoverflow question I cited no longer works. A particularly simple construction of a continuous function whose graph is everywhere locally of Hausdorff dimension $2$ is given in the following freely available paper: Peter Adolf Folke Wingren, Concerning a real-valued continuous function on the interval with graph of Hausdorff dimension $2$, L'Enseignement Mathématique (2) 41 #1-2 (January-June 1995), 103-110. – Dave L. Renfro Mar 31 '20 at 08:22
  • @Kavi Rama Murthy Thanks! – Rookie Mar 31 '20 at 09:36