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 &gt; 0, leading to a contradiction.

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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&#039;m trying to find a counter example for that but can&#039;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  &amp;gt;  0, leading to a contradiction.

Lamont Adkins · Accepted Answer

Take   A  =  {  0  }  ⊆      R  . For every open set   G containing   A there exists an   ε  &amp;gt;  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.

I`m trying to prove that: μ ( A ) = <munder> ∑

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