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Suppose that f is a measurable function $(\Omega, \mathfrak{F}, \mu)$ such that $\int_{A}f \, d\mu \geq 0 \forall A \in \mathfrak{F}$. Prove that $f \geq 0 \ \mu$-almost surely.

Hint: Let $A_n = (\omega \mid f(\omega) < \frac{-1}{n})$, and don't forget continuity.

Attempt:

$\forall n \in \mathbb{N}, A_n \subseteq A_{n+1}$ where $A_n$ is measurable since f is measurable (I think?).

By continuity of measure, lim$\mu(A_n)$ = $\mu(\bigcup_{n=1}^\infty A_n) $ where $\bigcup_{n=1}^{\infty} A_n = (\omega \mid f(\omega) < 0)$.

By monotonicity of measure, $0 \leq \mu(A_n) \leq \mu(A_{n+1})$.

Since we want to show that $\mu(\bigcup_{n=1}^\infty A_n) = 0$ (that is equivalent to $f \geq 0 \ \mu$-almost surely right?), and $(\mu(A_n))$ is a nonnegative increasing sequence, we must show that $\mu(A_n) = 0\ \forall n \in \mathbb{N}.$

I think the entire argument relies on the relationship between $\mu(A_n)$ and $\int_{A_n}f \, d\mu = \int_{\Omega}f*1_{A_n} \, d\mu$. What is it? Does nonnegativity of the integral mean that $\mu(A_n)= 0$? I am not quite sure that that necessarily follows.

BCLC
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1 Answers1

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Assume $\mu(A_n) > 0$. As $A_n \in \mathfrak{F}$, we then have that $\int_{A_n} f d\mu < -\frac{1}{n} \int_{A_n} d\mu = - \frac{1}{n} \mu(A_n) < 0$, which contradicts the assumption that $\int_A f d\mu \geq 0$ for all $A \in \mathfrak{F}$. Conclude that $\mu(A_n) = 0$ for any $n \in \mathbb{N}$.

msteve
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  • How do you know the first inequality holds? Thanks – BCLC Aug 23 '14 at 21:59
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    For any $\omega \in A_n$, $f(\omega) < -\frac{1}{n}$, and then integrate this expression over $A_n$. – msteve Aug 23 '14 at 22:00
  • msteve, you mean you just replaced f with $\frac{-1}{n}$ and the new integral is greater than the original integral by monotonicity of integration? thanks so much! – BCLC Aug 23 '14 at 22:07
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    Ya pretty much. Glad to help! – msteve Aug 23 '14 at 22:08
  • msteve, whenever I relate integrals and measure, is the trick then to compare the original integral with an integral of an indicator function? – BCLC Aug 23 '14 at 22:17
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    Often times that can work yes, though I would be hesitant to say that it is `the trick' of measure theory. – msteve Aug 23 '14 at 22:22