# Proof verification. The graph of a continuous function on a closed rectangle in $R^n$ is a null set in $R^{n+1}$.

Let $Q$ be a closed rectangle in $\mathbb{R}^n$ and let $f: Q \to \mathbb{R}$. The graph of $f$ is $G(f)=\{(x,y)\in \mathbb{R}^{n+1}\mid y=f(x)\}$. Show that if $f$ is continuous, then $G(f)$ has measure zero in $\mathbb{R}^{n+1}$.

I have written a proof of this, but I'd like to know if it is correct.

Proof. Since $Q$ is compact, $f$ is uniformly continuous on $Q$. So for any $\epsilon \gt 0$, we can find some $\delta \gt 0$ such that for all $x,y$ in some open rectangle inside $Q$ with norm less than $\delta$, we have $|f(x)-f(y)| \lt \epsilon/ 2v(Q)$, where $v(\cdot)$ is the volume of a rectangle. Now the collection of all such open rectangles cover $Q$, so by compactness, we can find $Q_1, \dots , Q_k$ that cover $Q$. Also note that the rectangles may intersect, but the intersection is also a rectangle, and so we can think of a collection of disjoint open rectangles $Q_1, \dots , Q_m$ whose union is $Q$. Then on each $Q_i$, we can form an interval $[a_i,b_i]$ where $a_i=\min_{Q_i} f, b_i=\max_{Q_i} f$. Then $(Q_i \times [a_i,b_i])_i$ covers $G(f)$ and $\sum v(Q_i \times [a_i,b_i])=\sum v(Q_i)\cdot v([a_i,b_i])\lt \epsilon/v(Q) \dot \sum v(Q_i)=\epsilon$. QED.

I'm not sure if I can form the disjoint rectangles as I have done. Is this proof correct? I would greatly appreciate any comments.

• Why is it fine if I overcount? Because then I may get $v(Q)\lt \sum v(Q_i)$, and this is why I chose a disjoint set. Commented Mar 2, 2016 at 22:03
• Sorry I don't follow what you mean. To show that it's a null set, we need to find a cover of rectangles that have volume less than $\epsilon$, but if we overcount, then we don't get that it's less than $\epsilon$. Commented Mar 3, 2016 at 0:36