Let X be a locally compact Hausdorff space and μ a Radon measure on X. If f , g : X → ( 0 , ∞ ) are continuous functions such that ∫ A f d μ = ∫ A g d μfor all Borel subsets A, can we deduce that f = g almost everywhere?(Note that in general f ( x ) ≠ g ( x ) for a point x ∈ X is possible, even if f, g are continuous, because a Radon measure may assign zero measure to an open subset of X!).My attempt:It is easy enough to see that f χ U = g χ U μ-almost everywhere for any U ⊆ X compact subset U. Indeed, by continuity and compactness f is bounded on U and compact subsets have finite measure. The same can be said about precompact open sets.However, X may not be σ-finite so I don't see how to deduce that f = g a.e.

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Let   X be a locally compact Hausdorff space and   μ a Radon measure on   X. If   f  ,  g  :  X  →  (  0  ,  ∞  ) are continuous functions such that      ∫    A    f  d  μ  =      ∫    A    g  d  μfor all Borel subsets   A, can we deduce that   f  =  g almost everywhere?(Note that in general   f  (  x  )  ≠  g  (  x  ) for a point   x  ∈  X is possible, even if   f,   g are continuous, because a Radon measure may assign zero measure to an open subset of   X!).My attempt:It is easy enough to see that   f      χ    U    =  g      χ    U     μ-almost everywhere for any   U  ⊆  X compact subset   U. Indeed, by continuity and compactness   f is bounded on   U and compact subsets have finite measure. The same can be said about precompact open sets.However,   X may not be   σ-finite so I don&#039;t see how to deduce that   f  =  g a.e.

Elianna Douglas · Accepted Answer

Take   A  =  {  x  :  f  (  x  )  &amp;gt;  g  (  x  )  }. If   μ  (  A  )  &amp;gt;  0 then there exists a compact set   K  ⊆  A such that   μ  (  K  )  &amp;gt;  0. Now       ∫                  K        N              f  d  μ  =      ∫                  K        N              g  d  μ where       K    N    =  {  x  ∈  K  :  f  (  x  )  ≤  N  ,  g  (  x  )  ≤  N  }. If   μ  (      K    N    )  &amp;gt;  0 then this is clearly a contradiction (since       ∫                  K        N              f  d  μ  &amp;gt;      ∫                  K        N              g  d  μ). Hence,   μ  (      K    N    )  =  0 for each   N which shows that   μ  (  A  )  =  0. Similarly,   μ  {  x  :  f  (  x  )  &amp;lt;  g  (  x  )  }  =  0 so   f  =  g a.e.Remark: Continuity of   f and   g is not used. Borel measurability is enough.

Let X be a locally compact Hausdorff space and μ a Radon measure on

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