Lebesgue measure proof for a set in R^2

  • Thread starter Thread starter rustyjoker
  • Start date Start date
  • Tags Tags
    Measure Proof Set
rustyjoker
Messages
5
Reaction score
0

Homework Statement



Prove (using Lebesgue outer measure) that $$m_{2}^{*}(\{(x,y)\in\mathbb{R}^{2}\colon x>1,0<y<x^{-2}\})=m_{2}^{*}(A)<\infty$$

The Attempt at a Solution



I'm not sure if this is valid proof but I'd have done it like this:
$$\int_{1}^{k}\left|x^{-2}\right|\, dx <\infty \forall k\in\mathbb{N} \Rightarrow \int_{[1,\infty[}\left|x^{-2}\right|\, dx<\infty \Rightarrow A $$is measurable and$$ m_{2}^{*}(A)=\int_{1}^{\infty}x^{-2}\, dx<\infty.\qquad\square$$What else should I include to make sure the proof is valid?
 
Physics news on Phys.org
rustyjoker said:
I'm not sure if this is valid proof but I'd have done it like this:
$$\int_{1}^{k}\left|x^{-2}\right|\, dx <\infty \forall k\in\mathbb{N} \Rightarrow \int_{[1,\infty[}\left|x^{-2}\right|\, dx<\infty$$
No, this implication is not valid. If you replace x^{-2} with x^{-1}, then your left hand inequality would still be true, but the right hand inequality would be false.

Similarly, the partial sums of an infinite series may all be finite, but that doesn't imply that the series converges. However, it is true that the series \sum_{n=1}^{\infty}n^{-2} converges. Can you use this series to bound your integral?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top