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Let $(\Omega,\Sigma)$ be a measurable space and $(\omega_k)_{k\in\mathbb{N}}$ a sequence of elements of $\Omega$. Let $$ \mathcal{M}:=\left\{\sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}: \quad(a_k)_{k\in\mathbb{N}}\subset(0,+\infty),\quad \sum_{k=1}^\infty a_k=1\right\}, $$ where $\delta_{\omega_k}$ denotes the point mass at $\omega_k$. Let $f:\Omega\rightarrow\mathbb{R}$ be a measurable function and denote by $\mathcal{N}$ the set of measures $\mu\in\mathcal{M}$ with respect to which $f$ is $\mu$-integrable. Assume $\mathcal{N}\neq\emptyset$. If $\vert f\vert$ is bounded, then we have the following implication: $$ \forall \mu\in\mathcal{N}\qquad\int_{\Omega} fd\mu\geq 0\Rightarrow \forall \omega\in\Omega:f(\omega)\geq 0.$$ But is it also necessary to have $\vert f\vert$ bounded, or can one weaken this condition?

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up vote 3 down vote accepted

If $|f|$ is not bounded, then $\int_{\Omega} f d \mu$ might not be well defined. Keep in mind that if $\mu= \sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}$ then

$$\int_{\Omega} f d \mu= \sum_{k=1}^\infty a_k\cdot f(\omega_k) \,.$$

Too see that you cannot drop the bounedndess, if $f$ is unbounded, for each $k$ pick some $\omega_k$ so that $|f(\omega_k)| > 2^k$, and set $\mu = \sum_{k=1}^\infty \frac{1}{2^{k+1}}\cdot\delta_{\omega_k}$.

Then $\int_{\Omega} f d \mu$ is an infinite series, each term being in absolute value at least one, and they can be positive and negative....

*P.S. * The above idea actually can be used to show that $f \in L^1(\mu) ; \forall \mu \in {\mathcal M} \Leftrightarrow |f|$ is bounded. Also, keep in mind that when you calculate the sum $ \sum_{k=1}^\infty a_k\cdot f(\omega_k) $ you must have absolute convergence, or "absolute" divergence to infinity, you cannot speak of conditional convergence because for the measure $\sum_{k=1}^\infty a_k\cdot\delta_{\omega_k}$, the order in which you list the terms doesn't matter.

The only way you could make sense of this by dropping the boundedness condition is by asking that for all $\mu \in {\mathcal M}$ you have either $f \in L^1(\mu)$ and $\int f d \mu \geq 0$ or $\int f d \mu = + \infty$. But this condition is equivalent to the negative part $f^-$ of $f$ is bounded....

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Thanks a lot for the comment. I have changed the question in that I am assuming that f is $\mu_0$-integrable for some $\mu_0\in\mathcal{M}$. I think under this assumption, the implication still holds for $\vert f\vert$ bounded, since one can construct a measure $\mu\in\mathcal{M}$, based on the measure $\mu_0$, which would give too much weight to the negative values of $f$ and make the integral negative... –  Andy Teich Jul 4 '12 at 15:41

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