# 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

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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.

Homework Helper
Gold Member
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.