# Graph of a continuous function has measure zero in $\mathbb{R}^2$

I've been working on the following exercise and can't quite get it. If anyone has any suggestions, please let me know.

Let $f \colon \mathbb{R} \to \mathbb{R}$ be continuous. Show that the graph of the function, $G_f = \{(x,f(x)) \colon x \in \mathbb{R}\}$ is a set of measure zero in $\mathbb{R}^2$.

My intuition: Consider writing $\mathbb{R} = \bigcup_{n \in \mathbb{Z} } [n,n+1]$, and note that since $f$ is continuous on $\mathbb{R}$, it's uniformly continuous on each compact interval comprising the union mentioned above. Thus, given any $\epsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < \delta$ for all $x,y \in [n,n+1]$.

I would like to use this idea, however, I'm not able to cook up a clean way to handle this when I union over all $n.$. :/

• Well, the countable union of measure-zero sets has measure zero. So, if you can handle the case of an interval, then you're done. – Omnomnomnom Sep 21 '16 at 3:03
• I didn't even think of that.. Thanks! – mathgenesis22813 Sep 21 '16 at 3:06

As you note, $$f:[0,1] \to \Bbb R$$ must be uniformly continuous. Fix an $$\epsilon > 0$$, and there exists a $$\delta$$ such that $$|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$$. If we chose $$n$$ so that $$1/n < \delta$$, then we can conclude that the graph of $$f$$ over $$[0,1]$$ is a subset of $$\bigcup_{k=0}^{n-1} [k/n,(k+1)/n] \times [f(k/n) - \epsilon, f(k/n) + \epsilon]$$ But then we have $$\mu\left[ \bigcup_{k=0}^{n-1} [k/n,(k+1)/n] \times [f(k/n) - \epsilon, f(k/n) + \epsilon] \right] \leq \\ \sum_{k=0}^{n-1} \mu([k/n,(k+1)/n] \times [f(k/n) - \epsilon, f(k/n) + \epsilon]) = n \cdot \frac{2 \epsilon}{n} = 2 \epsilon$$ Thus, we can conclude that the measure of the graph is less than $$2 \epsilon$$ for an arbitrary positive epsilon, which is to say that the measure must be zero.