We know when ( X , S , μ ) is a measure space, M R ( S ) consists signed (or real) measures on ( X , S ) is Banach space. The set V consists of the measures satisfying d v = h d μ for h ∈ L 1 ( μ ) is closed but how we can show that when μ is Lebesgue measure and S would be the cllections of Borel sets, the set V is not separable.

Question

We know when   (  X  ,  S  ,  μ  ) is a measure space,       M                  R              (  S  ) consists signed (or real) measures on   (  X  ,  S  ) is Banach space. The set V consists of the measures satisfying   d  v  =  h  d  μ for   h  ∈      L          1        (  μ  ) is closed but how we can show that when   μ is Lebesgue measure and S would be the cllections of Borel sets, the set V is not separable.

thoumToofwj · Accepted Answer

Denote   d      ν          a        =      χ          [      0      ,      a      ]        d  x for   a  &amp;gt;  0.The total variation is taken as a norm, assuming that   a  &amp;gt;  b, we compute that   ‖  d      ν          a        −  d      ν          b        ‖  =  ‖      χ          (      b      ,      a      ]        d  x  ‖  ≥  ∫      χ          [      0      ,      1      ]            χ          (      b      ,      a      ]        d  x  =  a  −  b.If we fix the   a and let   b  &amp;lt;  a  −  1, then   ‖  d      ν          a        −  d      ν          b        ‖  ≥  1. But there are uncountably many such   b, so   {  d      ν          b            }          b      &amp;lt;      a      −      1       cannot be approximated by a countable subset of       M                  R            .

We know when ( X , S , μ ) is a measure space, M <mrow

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