# Jordan measurable set

• Dec 23rd 2013, 05:40 AM
hedi
Jordan measurable set
http://www.tau.ac.il/~tsirel/Courses...sis3/lect6.pdf

Hi,
i need help with the solution of problem 6h3 (c),on page 87. in the attached link. for p not smaller than 1,i think i know,but there must be a nice and compact solution.
• Dec 23rd 2013, 10:28 AM
SlipEternal
Re: Jordan measurable set
Are you allowed to use results from topology in your proof? Define $\displaystyle d: \mathbb{R}^n \times \mathbb{R}^n \to [0,\infty )$ by

$\displaystyle d(x,y) = \sup_{1\le i\le n}|x_i - y_i|$

The topology this generates is frequently called the "Box metric". It is a routine exercise to show that these two metrics are topologically equivalent. A very similar proof can be used to show that each $\displaystyle E_p, p>0$ is Jordan measurable.
• Dec 23rd 2013, 11:56 AM
hedi
Re: Jordan measurable set
• Dec 23rd 2013, 02:33 PM
SlipEternal
Re: Jordan measurable set
Sorry, by two metrics, I meant the Euclidean metric and the Box metric are topologically equivalent. Those were the two metrics I meant. To show their topological equivalent, show that about every point of an open ball, there exists an open box centered at that point lying completely inside of the open ball. Then, show that about every point inside of an open box there exists an open ball centered at that point sitting inside of the box. (That is the topological equivalence proof). For Jordan measurability, you need to do something similar. Show that the $\displaystyle \sigma$-algebra generated by closed balls is equivalent to the $\displaystyle \sigma$-algebra generated by boxes. Since the set of Jordan measurable sets is a $\displaystyle \sigma$-algebra containing all boxes, it must contain the $\displaystyle \sigma$-algebra generated by closed balls, and the problem is solved.
• Dec 24th 2013, 12:59 PM
hedi
Re: Jordan measurable set
This holds for p not smaller than 1,when lp is a metric,right?
• Dec 24th 2013, 04:53 PM
SlipEternal
Re: Jordan measurable set
Quote:

Originally Posted by hedi
This holds for p not smaller than 1,when lp is a metric,right?

$\displaystyle E_0$ is just a point. It is trivially measurable (all sets in zero dimensions can be given measure zero). Any integration over zero dimensions is an integration over either zero or one point. So, the problem just doesn't make sense for p smaller than 1.
• Dec 24th 2013, 04:58 PM
hedi
Re: Jordan measurable set
so p must be an integer?
• Dec 24th 2013, 05:02 PM
SlipEternal
Re: Jordan measurable set
Quote:

Originally Posted by hedi
so p must be an integer?

Oh, I'm sorry. I misread the problem. p can be any real number greater than 0, so the problem should hold for p smaller than 1, as well. For p at least 1, it is an L_p space. You need to treat 0<p<1 separately.