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Consider the measure μ on R such that μ ( [ r , r ] ) > 0 for all r > 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?

Answer & Explanation



Beginner2022-06-26Added 19 answers

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].

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