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I'm now reading Lebesgue Integration on Euclidean Space of Frank Jones,but I don't have an idea of how to pave an irregular open set as in the problem below(Problem4(d) of Chapter2 in Lebegues Integration on Euclidean space,page 30):

In the plane $\mathbb{R}^{2}$ let $$ G=\left\{(x, y) \mid 1<x \text { and } 0<y<x^{-a}\right\} $$ where $a$ is a real number satisfying $a>1$. Prove that $\lambda(G)=\frac{1}{a-1}$.

I quote the definition of the measure of a open set in the book here:

$\lambda(G)=\sup \{\lambda(P) \mid P \subset G, P$ a special polygon $\}$

PhilippeMENG
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1 Answers1

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Possible approach. Take $$ P_i = \bigg[\frac{i}{n}, \frac{i + 1}{n}\bigg] \times \bigg[0, \bigg(\frac{i + 1}{n}\bigg)^{-a} \bigg] $$ Then $\bigcup_{n} P_i \subset G$ disjoint so we can try to calculate $$ \lambda\bigg(\bigcup_{i} P_i \bigg) = \sum_{i} \lambda(P_i) $$ using an appropriate Riemann sum. This approach is basically saying that $$ \lambda(G) = \int_{\mathbb{R}^2} \chi_{\{ x > 0\} \cap \{0 < y < x^{-a} \}}(x, y) d\lambda = \int_{1}^{\infty} x^{-a}dx = \frac{1}{a - 1} $$ where $\chi$ is the indicator function and $\{ x > 0\} = \{ (x, y) \in \mathbb{R}^2 : x > 1 \}$ and $\{0 < y < x^{-a} \} = \{ (x, y) \in \mathbb{R}^2 : 0 < y < x^{-a} \}$

Se7venn
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  • Thank you for the interpretation of Riemann approach in term of Lebesgue Integration.But the Riemman sum here is not easy to calculate and that's the main obstacle for me to deal with this problem. – PhilippeMENG Apr 12 '22 at 07:48
  • Per definition Riemann sum = some integral (i.e the one you want) so calculating it is straightforward. – Se7venn Apr 12 '22 at 16:35