Let f : X → R be a continuous function on a compact metric space X. Assume that a Borel probability measure μ is absolutely continuous with respect to Lebesgue measure Leb.Is it true that if f ( x ) &lt; 0 for Leb a.e. x, then ∫ f d μ &lt; 0 ??I think it should be true as μ &lt;&lt; Leb.Attempt: Assume that ∫ f d μ ≥ 0. Then f ≥ 0 μ a.e. x. That means μ ( { x : f ( x ) &gt; 0 } ) &gt; 0. That implies Leb ( ( { x : f ( x ) &gt; 0 } ) &gt; 0 which is not true.

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Let   f  :  X  →      R   be a continuous function on a compact metric space   X. Assume that a Borel probability measure   μ is absolutely continuous with respect to Lebesgue measure Leb.Is it true that if   f  (  x  )  &amp;lt;  0 for Leb a.e. x, then   ∫  f  d  μ  &amp;lt;  0  ??I think it should be true as   μ  &amp;lt;&amp;lt;  Leb.Attempt: Assume that   ∫  f  d  μ  ≥  0. Then   f  ≥  0 μ a.e. x. That means   μ  (  {  x  :  f  (  x  )  &amp;gt;  0  }  )  &amp;gt;  0. That implies   Leb  (  (  {  x  :  f  (  x  )  &amp;gt;  0  }  )  &amp;gt;  0 which is not true.

klemmepk · Accepted Answer

The compactness, metrizability and other conditions are unnecessary. All that is needed is that f is measurable and   μ  ≪  λ, where I use   λ for Lebesgue measure.  λ  {  x  ∈  X  :  f  (  x  )  ≥  0  }  =  0    ∧    μ  ≪  λ    ⟹    μ  {  x  ∈  X  :  f  (  x  )  ≥  0  }  =  0And immediately we have that, as X is not   μ-null:      ∫    X    f        d    μ  &amp;lt;  0I&#039;m afraid we do not quite have:      ∫    X    f        d    μ  ≥  0    ⟹    f  ≥  0    μ   - a.e.Since a function that is both positive and negative can simply cancel, e.g.      ∫                  R                        sin      ⁡      x        x          d    xFamously evaluates to   π  &amp;gt;  0, but of course             sin      ⁡      (      x      )        x   is negative on a set of infinite measure.Otherwise, what you said is fine.

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