Let $f\colon[a,b] \to \mathbb R$ be a continuous function. Show that its graph has measure zero.

I've tried with the following idea but I got stuck:

Let $\epsilon >0$, since $f$ is uniformly continuous, there exists $\delta>0$ such that $ |x-y|< \delta \implies |f(x)-f(y)|<\epsilon$. Let $P=\{x_0,\dots,x_n\}$ be a partition of the interval $[a,b]$ with $ |x_i-x_{i-1}|<\delta$.

The graph of $f$ is $G(f)=\{(x,f(x)) : x \in [a,b]\}$, and $G(f) \subset \bigcup_{i=1}^n [x_{i-1},x_i]\times[m_i,M_i]$, where $m_i=\min_{x \in [x_{i-1},x_i]} f(x)$,$M_i=\max_{x \in [x_{i-1},x_i]} f(x)$.

We have $ |G(f)|_e \leq \sum_{i=1}^nm([x_{i-1},x_i].m([m_i,M_i])<n\delta\epsilon$. At this point I got stuck, I need to arrive to an inequality with $\epsilon$ times a constant or something of the sort. I would appreciate some help.

  • $\begingroup$ $m([x_{i-1},x_i]$ means volume of $[x_{i-1},x_i]$, right? $\endgroup$
    – user295645
    Dec 4, 2019 at 1:26
  • 1
    $\begingroup$ @pozcukushimatostreet yes it’s the length of the interval. $\endgroup$
    – shalop
    Dec 4, 2019 at 1:50
  • $\begingroup$ $|G(f) |_e$ means measure of $G(f)$, right? $\endgroup$
    – user295645
    Dec 8, 2019 at 21:09
  • $\begingroup$ and how did you get '' $G(f) \subset \bigcup_{i=1}^n [x_{i-1},x_i]\times[m_i,M_i]$, where $m_i=\min_{x \in [x_{i-1},x_i]} f(x)$,$M_i=\max_{x \in [x_{i-1},x_i]} f(x)$ ''? $\endgroup$
    – user295645
    Dec 8, 2019 at 21:15

3 Answers 3


This isn't really relevant to this specific problem, but I just want to add that this result is still true for any Borel-measurable function. This is realized as a trivial consequence of Fubini's Theorem.

Let $m_1$ and $m_2$ denote 1- and 2-dimensional Lebesgue measure, respectively, and let $f: \mathbb{R} \to \mathbb{R}$ be measurable. Then $$ m_2(G_f) = \int_{\mathbb{R}^2} 1_{G_f} dm_2 = \int_{\mathbb{R}} \int_{\mathbb{R}} 1_{G_f}(x,y) dy dx = \int_{\mathbb{R}} m_1(\{ f(x) \}) dx = \int_{\mathbb{R}} 0 dx = 0$$

  • 4
    $\begingroup$ This is true, but one first needs to prove that $G_f$ is $m_2$-measurable. $\endgroup$ Mar 28, 2015 at 7:25
  • 6
    $\begingroup$ $G_f$ is always measurable (when $f$ is measurable), because $G_f=h^{-1}(\{0\})$, where $h: \mathbb{R}^2 \to \mathbb{R}$ is the measurable mapping defined by $h(x,y)=y-f(x)$. $\endgroup$
    – shalop
    Mar 28, 2015 at 7:33
  • $\begingroup$ And $h$ is the difference of two measurable functions $(x,y) \to y$ and $(x,y) \to f(x)$, and so it's measurable. $\endgroup$
    – Dzoooks
    Jul 8, 2019 at 17:38

HINT: your $n\delta$ can be replaced by $b-a$.

  • $\begingroup$ Oh, you're right! Is my solution ok with that correction? $\endgroup$
    – user156441
    Mar 28, 2015 at 5:29
  • 1
    $\begingroup$ @user156441 Yes. The uniform continuity is the main observation. $\endgroup$
    – Kola B.
    Mar 28, 2015 at 5:32
  • $\begingroup$ Since $x_i - x_{i-1} < \delta$ for all $i$, shouldn't $b - a$ be less than $n\delta$? $\endgroup$
    – kobe
    Mar 28, 2015 at 5:40
  • $\begingroup$ @kobe $x_i$'s form a partition. Only intervals of this partition are short. $\endgroup$
    – Kola B.
    Mar 28, 2015 at 5:42
  • 2
    $\begingroup$ @kobe Oh, I understand your point, I hope. But after the suggested correction we have not $n\delta$, but simply $|G(f)|_\epsilon<(b-a)\epsilon$. $\endgroup$
    – Kola B.
    Mar 28, 2015 at 5:57

Since $m([m_i,M_i]) < \varepsilon$ for all $i \in \{0,1,\ldots, n\}$,

$$\sum_{i = 1}^n m([x_{i-1},x_i])m([m_i,M_i]) < \sum_{i = 1}^n (x_i - x_{i-1})\varepsilon = (b - a)\varepsilon.$$


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