Lebesgue measure of $\{(x,y)\in\mathbb R^2:xy=1\}$ Let $A=\{(x,y)\in\mathbb R^2:xy=1\}$ and $\lambda$ be the Lebesgue measure on $\mathbb R^2$. Compute $\lambda(A)$.
I think $\lambda(A)=0$ because $A$ is the graph of the function $f(x)=\frac{1}{x}$ and the measure of the graph of a continuous function is  0. Is this true?
 A: $(1).$ Let $f:[a,b]:\to\Bbb R$ be continuous, where $a,b\in \Bbb R$ with $a\le b.$ Let $F$ be the graph of $f.$ (To a set-theorist, $F$ $is$ $f.$) We show $\lambda (F)\le 6r(b-a)$ for any $r\in\Bbb R^+$ and therefore $\lambda (F)=0$:
Take $n\in\Bbb Z^+$ large enough that $\forall x,y\in [a,b]\,(|x-y|<(b-a)/n\implies |f(x)-f(y)|<r).$ This is possible because $f$ is uniformly continuous on $[a,b].$
For convenience, for $j\in \Bbb Z$ let $x_j=a+j\cdot\frac {b-a}{2n}.$
Then $C=\{(x_j, x_{j+2})\times (f(x_{j+1})-r,f(x_{j+1})+r): -1\le j\le 2n-1\}$ is an open cover of $F$ and $\lambda (\cup C)\le (2n+1)\cdot\frac {b-a}{n}\cdot 2r=r(b-a)\frac {4n+2}{n}\le 6r(b-a).$
$(2)$. Let $F=\{(x,1/x):0\ne x\in \Bbb R\}.$ Let $\Bbb R=\cup D$ where $D$ is a countable family of closed bounded intervals. For example, let $D=\{[n,n+1]: n\in\Bbb Z\land -1\ne n\ne 0\}\cup \{[-1,-2^{-n}]:n\in\Bbb Z^+\}\cup \{[2^{-n},1]:n\in\Bbb Z^+\}.$
Let $D=\{D_m:m\in\Bbb Z^+\}.$ Let $F_m=\{(x,1/x): x\in D_m\}.$ By $(1)$ we have $\lambda (F_m)=0.$ So $\lambda (F)=\lambda (\cup_{m\in\Bbb Z^+}F_m)\le\sum_{m\in\Bbb Z^+}\lambda  (F_m)=0.$
A: This is a consequence of Fubini-Tonelli Theorem. Firstly, it is routine
to prove that $A$ is closed, and hence $A$ is a Borel set. Since
the usual topology on $\mathbb{R}$ is second countable, we have that
$\mathcal{B}(\mathbb{R}\times\mathbb{R})=\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
Therefore, $A\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
By Fubini-Tonelli Theorem, we have
$$
\lambda(A)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}1_{A}(x,y)dy\right)dx.
$$
For each $x\in\mathbb{R}\setminus\{0\}$, $1_{A}(x,y)=1$ iff $y=1/x$.
That is, $\{y\in\mathbb{R}\mid1_{A}(x,y)=1\}$ is a singleton and
hence $\int_{\mathbb{R}}1_{A}(x,y)dy=0$. It follows that $\lambda(A)=\int_{\mathbb{R}}0dx=0$.
