I'm proving this lemmaIf μ is a finite Borel measure on a metric space X and A a collection of mutually disjoint Borel sets of X, then at most countably many elements of A have nonzero μ-measure.This boils down to below result.Let ( a i ) i ∈ I be a collection of non-negative real numbers. Let's define ∑ i ∈ I a i := sup { ∑ i ∈ J a i | J is a countable subset of I } .If ∑ i ∈ I a i &lt; ∞, then at most countably many a i 's are positive.Could you have a check on my attempt?Let J := { i ∈ I ∣ a i &gt; 0 } and J r := { i ∈ I ∣ a i ≥ r } for r ∈ Q &gt; 0 . Then J = ⋃ r ∈ Q &gt; 0 J r .It follows from ∑ i ∈ I a i &lt; ∞ that J r is countable for all r ∈ Q &gt; 0 . Then it follows from Q &gt; 0 is countable that J is countable.

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I&#039;m proving this lemmaIf   μ is a finite Borel measure on a metric space   X and       A   a collection of mutually disjoint Borel sets of   X, then at most countably many elements of       A   have nonzero   μ-measure.This boils down to below result.Let   (      a    i        )          i      ∈      I       be a collection of non-negative real numbers. Let&#039;s define      ∑          i      ∈      I            a    i    :=  sup      {          ∑              i        ∈        J                    a      i                    |                J     is a countable subset of     I    }    .If       ∑          i      ∈      I            a    i    &amp;lt;  ∞, then at most countably many       a    i  &#039;s are positive.Could you have a check on my attempt?Let   J  :=  {  i  ∈  I  ∣      a    i    &amp;gt;  0  } and       J    r    :=  {  i  ∈  I  ∣      a    i    ≥  r  } for   r  ∈            Q              &amp;gt;      0      . Then  J  =      ⋃          r      ∈                        Q                          &amp;gt;          0                          J    r    .It follows from       ∑          i      ∈      I            a    i    &amp;lt;  ∞ that       J    r   is countable for all   r  ∈            Q              &amp;gt;      0      . Then it follows from             Q              &amp;gt;      0       is countable that   J is countable.

Schetterai · Accepted Answer

The proof is correct.Mind, each       J    r   is not only countable, but actually finite.

I'm proving this lemma If μ is a finite Borel measure on a metric space

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