Given a Borel set B of finite Lebesgue measure show that for every &#x03F5;<!-- ϵ --> &

Wade Bullock

Wade Bullock

Answered question

2022-07-05

Given a Borel set B of finite Lebesgue measure show that for every ϵ > 0, there exist: a compact set C 1 B, a closed set C 2 R B and a continuous function ϕ : R [ 0 , 1 ] s.t.:
χ C 1 ϕ χ R C 2
| | χ B ϕ | | 1 < ϵ"

Soo I have only the vague intuition of using the monotone convergence theorem somehow - e.g. perhaps setting B [ n , n + 1 ] as sets on which corresponding simple functions f n (such as the indicator function) can be defined so that they are wedged between χ C 1 and χ R C 2 (thus satisfying the first condition) and so that taking the limit on their integral would satisfy the second condition.
I am not sure how to go about accomplishing this, especially since the example I suggested for B [ n , n + 1 ] is not necessarily closed and therefore not necessarily compact. In addition, I am not sure how to define the required closed set C 2 .
Any guidance on the steps of applying MCT (assuming I am at least right on this) to this problem would be much appreciated.
Edit: Another thought I had was perhaps using somehow the regularity of the Lebesgue measure to define the compact and closed sets (?), i.e. there are a compact set K and an open set U such that:
μ ( B ) μ ( K ) + ϵ
and
μ ( B ) μ ( U ) ϵ
Maybe these could be used to define the indicator function I mentioned above?

Answer & Explanation

esperoanow

esperoanow

Beginner2022-07-06Added 11 answers

The problem is stated a bit weirdly. It is more commonly stated with U = R C 2 . To solve your problem, you can use regularity of Lebesgue measure to get K and U, and then use Urysohn's lemma to get ϕ C c ( R ) (or even in C c ( R )) with K ϕ U.

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