The ingerals of a function on two different measures

[zyp]

Is the integral of a strictly positive function on a set of positive measure strictly positive? Thankis a lot

 

[mathman]

Yes. I don't know how to elaborate, but if you try to prove otherwise (integral = 0), you will get a contradiction.

 

[quasar987]

Yes. Suppose f:A-->R is such a function. We have that

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

and so by subadditivity of the measure

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

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

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

for some n. By definition

\int_Af=\sup_h\left{\int_Ah\right}

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

It follows that

\int_A f\geq \int_{A_n}\frac{1}{n}=\frac{1}{n}\mbox{mes}(A_n)>0

 

[zyp]

Yes. Suppose f:A-->R is such a function. We have that

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

and so by subadditivity of the measure

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

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

\int_Af=\sup_h\left{\int_Ah\right}

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

It follows that

\int_A f\geq \int_{A_n}\frac{1}{n}=\frac{1}{n}\mbox{mes}A_n>0

i see. thanks a lot indeed.