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I know that a random variable $X$ is integrable, that is $\mathbb{E}\left[|X| \right]<+\infty$. Can I apply the dominated convergence theorem and state that $\lim_{a \to +\infty} \mathbb{E} (|X|\ 1_{|X| \geq a})$ is equal to $\mathbb{E} \left[\lim_{a \to +\infty} |X| 1_{|X| \geq a}\right]$ and therefore equal to $0$ ?

Thank you very much for any help you can provide ! :)

Fred G.
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1 Answers1

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By definition a random variable $X$ is just a measurable function, hence it does not need to be integrable. Anyway, if you assume that $X$ is integrable, then (by DCT) $$ \lim_n \int f_n \mathrm{d}P = \int \lim_n f_n \mathrm{d} P. $$ whenever $f_n \le |X|$. Then, it is enough to choose $f_n=X \cdot {1}_{[n,\infty)}(|X|)$.

Paolo Leonetti
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  • Thank you !! By "I know that" I meant "by assumption", indeed :) Ok, so I can swap them without problems ! Have a nice day – Fred G. Nov 06 '15 at 09:59