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Consider the probability space $((0,1],\mathcal{B}((0,1]),\lambda|_{(0,1]})$ and define $$X_n(\omega) := \frac{1}{\omega} 1_{\big(0, \frac{1}{n}\big]}(\omega).$$

I am told that this R.V. is not integrable. Why is this the case? I know that for a R.V. to be integrable in$L^1$, it must have finite expectation. But I can not see why $E[X_n(w)]=\infty$.

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    $\begingroup$ Because $\int_0^{1\over n} {1 \over t} dt = \infty$. $\endgroup$ – copper.hat Dec 5 '19 at 20:53
  • $\begingroup$ So the lebesegue measure us actually $dP(w)=dw$? $\endgroup$ – Alchemy Dec 5 '19 at 20:54
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    $\begingroup$ Hmmmm, yes. What did you think it was? $\endgroup$ – copper.hat Dec 5 '19 at 20:55
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By definition, our probability measure here is $\lambda\vert_{(0,1]}$ so $\mathsf d\mathbb P(\omega) = \mathsf d\omega$ and $$ \mathbb E[X] = \int_\Omega X(\omega)\ \mathsf d\mathbb P(\omega) = \int_{(0,1/n]} \frac1\omega\ \mathsf d\omega = +\infty, $$ so $X$ is not integrable.

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