1. ## 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 = and λ(F)>1-ε, where λ is the Lebesgue measure.

Thank you very much!

2. 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?

3. I have seen this problem as an exercise in two books, then I think there is a solution.

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

The proof of this is easy: There exists a lower semicontinous function $f\geq 1_{X}$ with $\int_{\mathbb{R} ^n} f <\frac{ \varepsilon}{2}$ then take $\Omega := f^{-1} \left( \frac{1}{2} , \infty \right)$

5. Nice thinking 'Jose27'