Extract from a proof in a measure theory script:For each set A ⊂ R n and each r &gt; 0, we set. A r := { x ∈ A : B ( x , r ) ⊆ A }Let Σ be an algebra in R n containing all open subsets of R n , then for any set A ∈ Σ and any r &gt; 0, we have A r ∈ Σ.

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Extract from a proof in a measure theory script:For each set   A  ⊂            R              n       and each   r  &amp;gt;  0, we set.      A          r        :=  {  x  ∈  A  :  B  (  x  ,  r  )  ⊆  A  }Let   Σ be an algebra in             R              n       containing all open subsets of             R              n      , then for any set   A  ∈  Σ and any   r  &amp;gt;  0, we have       A          r        ∈  Σ.

Erick Blake · Accepted Answer

Let   d be the Euclidean metric on             R        n  . It is not hard to verify that      A    r    =  {  x  ∈            R        n    :  d  (  x  ,      A    c    )  ≥  r  }  .Since the map   f  :            R        n    →      R   defined by   f  (  x  )  =  d  (  x  ,      A    c    ) is continuous and       A    r    =      f          −      1        (  [  r  ,  ∞  )  ), it follows that       A    r   is closed. Since your algebra contains all open sets, it contains all complements of open sets, i.e. all closed sets.

bs1tuaz · Accepted Answer

My attempt to prove this statement: Let   U be the union of all balls       B    r    (  x  ) with   x  ∈      A    c    . Then       A    r    =  A  ∖  U  . It follows  x  ∈  U  ⇔  ∃  y  ∈      A    c    :  x  ∈      B    r    (  y  )    ⇔  ∃  y  ∈      A    c    :  y  ∈      B    r    (  x  )  ⇔      B    r    (  x  )  ⊄  AIt follows that  x  ∈  A  ∖  U  ⇔  x  ∈  A  ∧      B    r    (  x  )  ⊂  A  ⇔  x  ∈      A    r    .And thus       A    r    =  A  ∖  U  ∈  Σ  .. We know that   U  ∈  Σ since   U is a union of open sets.

Extract from a proof in a measure theory script: For each set A ⊂ <mro

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