Let X be a set, A a ring of subsets of X, μ ∶ A → R ¯ ≥ 0 a premeasure and μ ∗ the outer measure generated by μ. (By Caratheodory)If E ∈ M μ ∗ and satisfacies μ ∗ ( E ) &lt; ∞, then for each ε existx A ∈ A such that μ ∗ ( A △ E ) &lt; εI try to test this, my first idea was to give a cover of E, I just don't know if I can find said cover in A , as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed?

Question

Let   X be a set,       A   a ring of subsets of   X,   μ  ∶      A    →                    R            ¯              ≥      0       a premeasure and       μ    ∗   the outer measure generated by   μ. (By Caratheodory)If   E  ∈            M                      μ        ∗             and satisfacies       μ    ∗    (  E  )  &amp;lt;  ∞, then for each   ε existx   A  ∈      A   such that       μ    ∗    (  A  △  E  )  &amp;lt;  εI try to test this, my first idea was to give a cover of   E, I just don&#039;t know if I can find said cover in       A  , as the measure is not finite, so the extension is not unique, someone can give me a hint how to proceed?

iskakanjulc · Accepted Answer

By definition of the outer measure, there is a countable collection   {      Q    j    } of elements in       A   such that      ∑          j      =      1              ∞        μ  (      Q    j    )  ≤      μ    ∗    (  E  )  +  ε  .Since       μ    ∗    (  E  )  &amp;lt;  ∞, the series above converges. Hence there is an   N so that      ∑          j      =      N      +      1              ∞        μ  (      Q    j    )  &amp;lt;      ε    2    .Consider   A  =      ⋃          j      =      1        N        Q    j  .

Addison Trujillo · Accepted Answer

Be careful, since       μ    ∗    (  E  )  =  inf  {      ∑          j      =      1        ∞    μ  (      Q    j    )  :  {      Q    j        }          j      =      1        ∞    ⊆      A    ,  E  ⊆      ⋃          j      =      1        ∞        Q    j    } is not   ∞,       μ    ∗    (  E  )  +  ε  (  &amp;gt;      μ    ∗    (  E  )  ) is not an lower bound for   {      ∑          j      =      1        ∞    μ  (      Q    j    )  :  {      Q    j        }          j      =      1        ∞    ⊆      A    ,  E  ⊆      ⋃          j      =      1        ∞        Q    j    }, and so there exists   {      Q    j        }          j      =      1        ∞    ⊆      A   with   E  ⊆      ⋃          j      =      1        ∞        Q    j   such that       ∑          j      =      1        ∞    μ  (      Q    j    )  &amp;lt;      μ    ∗    (  E  )  +  ε. In other words, the condition       μ    ∗    (  E  )  &amp;lt;  ∞ is necessary for this step.