I`m trying to prove that: μ ( A ) = ∑ n ∈ N δ...

fabios3

fabios3

Answered

2022-06-27

I`m trying to prove that:
μ ( A ) = n N δ 1 n ( A )
is not a regular measure.
A measure is regular if:
μ ( A ) = inf { μ ( G ) | A G , G  is open } = sup { μ ( F ) | F A , F  is closed }
for any measurable A.
I'm trying to find a counter example for that but can't come up with any.
I was thinking about a set which contains a finite number of 1/n but can only be covered by ( 0 , ϵ ) for some ϵ > 0, leading to a contradiction.

Answer & Explanation

Lamont Adkins

Lamont Adkins

Expert

2022-06-28Added 11 answers

Take A = { 0 } R . For every open set G containing A there exists an ε > 0 such that ] ε , ε [ G. This open interval contains infinitely many 1 n . Thus μ ( G ) μ ( ] ε , ε [ ) = . But μ ( A ) = 0. This shows that μ is not outer regular.

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