# Proving a function is not integrable

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?

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$$\int_X f d \mu > \sum_{k = 1}^\infty \frac{k}{k+1} = \infty$$ –  Matt N. Aug 20 '12 at 14:49

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

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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.