# Is the following rv integrable?

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 $$E[X_n(w)]=\infty$$.

• Because $\int_0^{1\over n} {1 \over t} dt = \infty$. – copper.hat Dec 5 '19 at 20:53
• So the lebesegue measure us actually $dP(w)=dw$? – Alchemy Dec 5 '19 at 20:54
• Hmmmm, yes. What did you think it was? – copper.hat Dec 5 '19 at 20:55

## 1 Answer

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.