Consider the measure μ on R such that μ ( [ − r , r ] ) &gt; 0 for all r &gt; 0.Can we construct a (smooth) function f satisfying μ − f d x ≥ 0 in a measure sense? If μ has a continuous density g, then it seems easy. But what conditions are needed for the existence of f for a measure μ? or is it possible always?

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Consider the measure   μ on   R such that   μ  (  [  −  r  ,  r  ]  )  &amp;gt;  0 for all   r  &amp;gt;  0.Can we construct a (smooth) function   f satisfying   μ  −  f  d  x  ≥  0 in a measure sense? If   μ has a continuous density   g, then it seems easy. But what conditions are needed for the existence of   f for a measure   μ? or is it possible always?

popman14ee · Accepted Answer

I suppose you want   f to be a non-negative smooth function.If   (      r    n    ) is an evaluation of rational numbers and   μ  =  ∑      1          2              n                  δ                  r        n             then the only function   f satsfying your conditioin is   f=0 (up to a Lebesgue null set). This becasue       ∫    E    f  (  x  )  d  x  ≤  m  (  E  )  =  0 for any Borel set   E contained in       R    ∖      Q   and this implies       ∫    E    f  =  0 for all Borel sets   E.Note: If a smooth function is 0 a.e. [Lebesgue] then it is 0 at every point.The same conclusin holds for any   μ singualr w.r.t. Lebesgue measure. If   μ is concentared on   S and   m (Lebesgue measure) on it complement then   μ  (  E  )  ≥      ∫    E    f  =  0 for all   E in side       S          c       and       ∫    E    f  (  x  )  x  d  =  0 for   E inside   S also. So   f  =  0 a.e. [Lebesgue].

Consider the measure μ on R such that μ ( [

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