I'm reading page 30 on Lebesgue integration on Euclidean space by Frank Jones.
I only have definition $\lambda(G)=sup\{\lambda(P) | P\subset G \}$ where P is special polygon defined by union of the form $\cup_N [a_1,b_1] \times [a_n,b_n]$. and what I have to prove is :
$\lambda(G)=1/(a-1)$ where $G=\{(x,y)|1<x \wedge 0<y<x^{-a} \}$ with $a>1$
I tried this by splitting above set like when we split curve by small regular rectangles. but I failed to find the value since terms are so complicated :
let $$P^k_n:=\cup_{1\leq i<n}[\frac{i}{n},\frac{i+1}{n}]\times[\frac{1}{n^2 (k+ \frac{i+1}{n})^a},\frac{1}{(k+ \frac{i+1}{n})^a}-\frac{1}{n^2 (k+ \frac{i+1}{n})^a}]$$. then as $n$ goes to infinity, we can expect that right side of product goes to $[0,k^a]$.
I expressed $\lambda(P^k_n)=\frac{1}{n} \cdot (\frac{1}{(k+ \frac{2}{n})^a}+\frac{1}{(k+ \frac{3}{n})^a}+\cdots+\frac{1}{(k+ \frac{n}{n})^a}-a_n)$ where $\lim_{n\to \infty} a_n=0$. This equals $$\frac{1}{n} \Bigg( \frac{\displaystyle\prod_{1\leq i<n,d\ i\neq 1} (k+\frac{i+1}{n})}{(k+ \frac{2}{n})\cdots(k+\frac{n}{n})}\Bigg)^a+\cdots +\frac{1}{n}\Bigg( \frac{\displaystyle\prod_{1\leq i<n,d\ i\neq n-1} (k+\frac{i+1}{n})}{(k+ \frac{2}{n})\cdots(k+\frac{n}{n})}\Bigg)^a$$ If fix $k$ and consider only last terms then I can think of this as $$\frac{1}{n}[(\frac{\frac{n!/2}{n^{n-2}}}{n!/n^{n-1}})^a+(\frac{\frac{n!/3}{n^{n-2}}}{n!/n^{n-1}})^a+(\frac{\frac{n!/n}{n^{n-2}}}{n!/n^{n-1}})^a]=\frac{1}{n}[(\frac{n}{2})^a+(\frac{n}{3})^a+\cdots (\frac{n}{n})^a]$$ I couldn't do further since I didn't know what is the next step and there's no $k$ terms which should be used later. I tried other terms but I don't see proper terms
The area of the union of the rectangles of width $\frac1n$ below the upper boundary of $G$ is $$A_n=\sum_{k=n}^\infty\frac1n\cdot\frac1{((k+1)/n)^a}=n^{a-1}\sum_{k=n+1}^\infty\frac1{k^a}$$ To estimate the RHS, note that, for every $k\geqslant2$, $$\frac1{k^{a-1}}-\frac1{(k+1)^{a-1}}\leqslant\frac{a-1}{k^a}\leqslant\frac1{(k-1)^{a-1}}-\frac1{k^{a-1}}\tag{$\ast$}$$ hence $$n^{a-1}\sum_{k=n+1}^\infty\left(\frac1{k^{a-1}}-\frac1{(k+1)^{a-1}}\right)\leqslant(a-1)A_n\leqslant n^{a-1}\sum_{k=n+1}^\infty\left(\frac1{(k-1)^{a-1}}-\frac1{k^{a-1}}\right)$$ that is, $$\left(\frac{n}{n+1}\right)^{a-1}\leqslant(a-1)A_n\leqslant1$$ from which the limit of $A_n$ follows.
To show $(\ast)$, note that $(\ast)$ is implied by $$1+bx\leqslant(1-x)^{-b}\tag{$\ast\ast$}$$ for $$b=a-1\qquad x=\pm\frac1k$$ and that $(\ast\ast)$ is direct by convexity of its RHS.