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Let $f: \Omega \to \mathbb{R}$ be a measurable function defined on a canonical probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Here, we can set $\Omega := \mathbb{R}^n$, and the measurability always refers to Borel measurability.

It is well-know that if $f \geq 0$ $\mathbb{P}$-a.e., then $\int_\Omega f (x) \, \mathbb{P}(\mathrm{d} x) \geq 0$.

I am curious that could we get a stricter result such that $\int_\Omega f (x) \, \mathbb{P}(\mathrm{d} x) > 0$ whenever $f(x) > 0$ for each $ x \in \Omega$?

The answer seems to be YES, but I am not sure how to prove it rigorously. Could anyone help me out please?

My attempt:

I proved the case where $f$ is continuous (rather than just measurable) on $\Omega$. To see this, pick any compact subset $K \subset \Omega$, and then $f(x) \geq \eta= \max_{x \in K} f(x) >0$ on $K$.

It then follows that $\int_{\Omega} f(x) \, \mathbb{P}(\mathrm{d} x) \geq \int_{K} f(x) \, \mathbb{P}(\mathrm{d} x) \geq \int_{K} \eta \, \mathbb{P}(\mathrm{d} x) = \eta >0.$

However, I did not know how to show such result for the case of a measurable function $f$ on $\Omega$.

Would you mind giving me some guidance or hint on it please? Thank you so much in advance!

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If $f(x)>0$ for all $x$, then for some positive integer $n$ the set $A_n=\{x:f(x)>1/n\}$ has positive measure (since $\bigcup_n A_n=\Omega$). We thus have $$\int_\Omega f dP\geq \int_{A_n} f dP\geq \int_{A_n} \frac{1}{n}dP=\frac{P(A_n)}{n}>0.$$

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  • $\begingroup$ Thank you very much @Eric Wofsey. $\endgroup$ – Paradiesvogel May 29 '18 at 1:32

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