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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.

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In fact, making the rectangles disjoint is neither hard nor necessary; there's no problem in over-counting volumes as your inequality will still hold.

Meanwhile, if you want disjoint rectangles you could just subdivide them and count the intersection just once. Since there are only finitely many rectangles, it's fine to think about how to do this for just two rectangles.

Good proofing and take care.

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  • $\begingroup$ 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. $\endgroup$ – nomadicmathematician Mar 2 '16 at 22:03
  • $\begingroup$ The measure of the graph is bounded by the summed measure of the disjoint rectangles you propose, so it continues to be bounded by something strictly larger (double counting volumes). $\endgroup$ – D. Wagner Mar 2 '16 at 23:38
  • $\begingroup$ 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 $\epsilon$. $\endgroup$ – nomadicmathematician Mar 3 '16 at 0:36

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