Let X be any locally compact Hausdorff space, and μ be a positive finite regular (both inner and outer) measure on X such that supp ⁡ ( μ ) is compact. We have that ∫ X f d μ = 0for each f ∈ C c ( X ).Then can we conclude that μ = 0?I was trying to find a counter-example, but am stuck at some technical issue.

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Let   X be any locally compact Hausdorff space, and   μ be a positive finite regular (both inner and outer) measure on   X such that   supp  ⁡  (  μ  ) is compact. We have that      ∫    X    f     d  μ  =  0for each   f  ∈      C    c    (  X  ).Then can we conclude that   μ  =  0?I was trying to find a counter-example, but am stuck at some technical issue.

Anika Stevenson · Accepted Answer

Yes, you may.Note that       I    μ    :      C    0    (  X  )  →      R   given by       I    μ    (  f  )  =      ∫    X    f  d  μ is a continuous linear functional. Since       C    c    (  X  ) is dense in       C    0    (  X  ) and       I    μ   is zero on a dense set, it must extend uniquely to the zero functional on space       C    0    (  X  ). But the zero functional is represented by integration against the zero measure, so   μ  =  0.

Let X be any locally compact Hausdorff space, and μ be a positive finite re

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