The ingerals of a function on two different measures

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  • #1
zyp
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Is the integral of a strictly positive function on a set of positive measure strictly positive? Thankis a lot
 
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  • #2
mathman
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Yes. I don't know how to elaborate, but if you try to prove otherwise (integral = 0), you will get a contradiction.
 
  • #3
quasar987
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Yes. Suppose f:A-->R is such a function. We have that

[tex]A=\bigcup_{n\in\mathbb{N}}f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right)[/tex]

and so by subadditivity of the measure

[tex]\mbox{mes}(A)\leq \sum_n\mbox{mes}\left(f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right)\right)[/tex]

Since mes(A)>0, it must be that

[tex]A_n:=\mbox{mes}\left(f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right)\right)>0[/tex]

for some n. By definition

[tex]\int_Af=\sup_h\left{\int_Ah\right}[/tex]

where h:A-->R denotes a positive measurable stair function bounded above by f.

It follows that

[tex]\int_A f\geq \int_{A_n}\frac{1}{n}=\frac{1}{n}\mbox{mes}(A_n)>0[/tex]
 
  • #4
zyp
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Yes. Suppose f:A-->R is such a function. We have that

[tex]A=\bigcup_{n\in\mathbb{N}}f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right)[/tex]

and so by subadditivity of the measure

[tex]\mbox{mes}(A)\leq \sum_n\mbox{mes}\left(f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right)\right)[/tex]

Since mes(A)>0, it must be that A_n:=mes(f^{-1}\left(\left[\frac{1}{n},+\infty\right)\right))>0 for some n. By definition

[tex]\int_Af=\sup_h\left{\int_Ah\right}[/tex]

where h:A-->R denotes a positive measurable stair function bounded above by f.

It follows that

[tex]\int_A f\geq \int_{A_n}\frac{1}{n}=\frac{1}{n}\mbox{mes}A_n>0[/tex]
i see. thanks a lot indeed.
 
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