# Measure

• Jun 19th 2010, 06:51 PM
eltondelamancha
Measure
I'm having trouble, please can somebody help me?

Prove that for all ε>0 there exists a closed set Fc[0,1] such that F http://upload.wikimedia.org/math/e/7...5a73d4e6b8.png http://upload.wikimedia.org/math/2/b...3d4231ab04.png = http://upload.wikimedia.org/math/d/0...709b02e52e.png and λ(F)>1-ε, where λ is the Lebesgue measure.

Thank you very much!
• Jun 19th 2010, 07:14 PM
Ackbeet
That sounds like a very difficult problem, if not impossible. The irrationals in the interval [0,1] are not closed, and I'm not sure I see how any subset of them can be closed. Are you sure this problem is solvable?
• Jun 20th 2010, 07:57 AM
eltondelamancha
I have seen this problem as an exercise in two books, then I think there is a solution.
• Jun 21st 2010, 11:38 AM
Jose27
Quote:

Originally Posted by eltondelamancha
I'm having trouble, please can somebody help me?

Prove that for all ε>0 there exists a closed set Fc[0,1] such that F http://upload.wikimedia.org/math/e/7...5a73d4e6b8.png http://upload.wikimedia.org/math/2/b...3d4231ab04.png = http://upload.wikimedia.org/math/d/0...709b02e52e.png and λ(F)>1-ε, where λ is the Lebesgue measure.

Thank you very much!

This is an immediate consequence of the following lemma:

A set $\displaystyle X\subset \mathbb{R} ^n$ has measure $\displaystyle 0$ if and only if for all $\displaystyle \varepsilon >0$ there exists an open set $\displaystyle X\subset \Omega$ with $\displaystyle \lambda (\Omega ) <\varepsilon$

The proof of this is easy: There exists a lower semicontinous function $\displaystyle f\geq 1_{X}$ with $\displaystyle \int_{\mathbb{R} ^n} f <\frac{ \varepsilon}{2}$ then take $\displaystyle \Omega := f^{-1} \left( \frac{1}{2} , \infty \right)$
• Jun 25th 2010, 08:29 PM
Idealconvergence
Nice thinking 'Jose27'