Preamble: Suppose that m ∗ is the Radon outer measure on R n and S ⊆ R n . Define r i ∗ ( S ) = ( m ∗ | B i ( 0 ) ) ( S ) = m ∗ ( S ∩ B i ( 0 ) ) , S ⊆ R n the restriction of m ∗ to the open ball of radius i centered at the origin. Assume it to be known that 1.) r i ∗ is a Radon outer measure, and r i ∗ ( R n ) ≤ m ∗ ( B i ( 0 ) ¯ ) &lt; ∞, 2.) for every ϵ &gt; 0 there exists closed set C and open set D such that C ⊆ S ⊆ D, m ∗ ( S ∖ C ) &lt; ϵ , m ∗ ( D ∖ S ) &lt; ϵ, 3.) if the set S is m ∗ -measurable (by Caratheodory's criterion), then it is also r i ∗ -measurable.Then, there exists open sets S ⊆ D i such that r i ∗ ( D i ∖ S ) &lt; ϵ 2 i + 1 , i = 1 , 2 , …. Define D = ⋃ i = 1 ∞ ( D i ∩ B i ( 0 ) ).Main question: What I am struggling to understand is that why S ⊂ D? We certainly know that S belongs to every D i . But what "guarantee" there is that S belongs to the intersection of D i with the open ball B i ( 0 )? My reading material does not provide any further information about this, and proceeds onwards with the surrounding proof about a property of Radon outer measures.

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Preamble: Suppose that       m    ∗   is the Radon outer measure on             R        n   and   S  ⊆            R        n  . Define       r    i    ∗    (  S  )  =  (      m    ∗        |        B    i    (  0  )  )  (  S  )  =      m    ∗    (  S  ∩      B    i    (  0  )  )  ,  S  ⊆            R        n   the restriction of       m    ∗   to the open ball of radius i centered at the origin. Assume it to be known that 1.)       r    i    ∗   is a Radon outer measure, and       r    i    ∗    (            R        n    )  ≤      m    ∗    (                    B        i            (      0      )        ¯    )  &amp;lt;  ∞, 2.) for every   ϵ  &amp;gt;  0 there exists closed set C and open set D such that   C  ⊆  S  ⊆  D,       m    ∗    (  S  ∖  C  )  &amp;lt;  ϵ  ,      m    ∗    (  D  ∖  S  )  &amp;lt;  ϵ, 3.) if the set S is       m    ∗  -measurable (by Caratheodory&#039;s criterion), then it is also       r    i    ∗  -measurable.Then, there exists open sets   S  ⊆      D    i   such that       r    i    ∗    (      D    i    ∖  S  )  &amp;lt;      ϵ          2              i        +        1            ,   i  =  1  ,  2  ,  …. Define   D  =      ⋃          i      =      1        ∞    (      D    i    ∩      B    i    (  0  )  ).Main question: What I am struggling to understand is that why   S  ⊂  D? We certainly know that S belongs to every       D    i  . But what &quot;guarantee&quot; there is that S belongs to the intersection of       D    i   with the open ball       B    i    (  0  )? My reading material does not provide any further information about this, and proceeds onwards with the surrounding proof about a property of Radon outer measures.

Jaylee Dodson · Accepted Answer

Take   x  ∈  S and find   i so large that       |    x      |    &amp;lt;  i. Then   x  ∈      D    i    ∩  B  (  0  ,  i  ).

Preamble: Suppose that m ∗ </msup> is the Radon outer measure o

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