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Let $(X, \Sigma,\mu)$ be a $\sigma$-finite measure space, and let $f:X\to\mathbb R^+$ be a measurable function such that $$\mu(\{x\in X\,\colon\,f(x)>t\})>\frac{1}{1+t},\; \forall t>0.$$ Prove then that $f$ is not integrable.

I've tried to derive a contradiction, however my original plan to use Chebyshev inequality to get such an absurd turned out to be useless..

Can you help me?

Thanks in advance, Guido

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$$ \int_X f d \mu > \sum_{k = 1}^\infty \frac{k}{k+1} = \infty$$ – Rudy the Reindeer Aug 20 '12 at 14:49
up vote 4 down vote accepted

$$\int_X f d\mu = \int_\mathbb{R_+}\mu(\{x: f(x)>t \})dt > \int_0^\infty \frac{1}{1+t}$$

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If $f$ is measurable, then $$ \int_X f \, d\mu = \int_0^{+\infty} \mu \left( \left\{ x \in X \mid f(x)>t \right\} \right)\, dt $$ See this link.

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Hey Siminore! This time I beat you by 30 seconds! And I have an additional term! – user20266 Aug 20 '12 at 14:51
+1 to your question and to your speed :-) – Siminore Aug 20 '12 at 14:57
Which question? +1, cause you gave the same answer ;-) – user20266 Aug 20 '12 at 15:17
Sorry, +1 to your answer :-) – Siminore Aug 20 '12 at 15:37

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