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Let $X$ be a random variable and $(X_n)_{n\geq 0}$ be a sequence of random variables.

Show that $\lim_{n\to\infty}X_n=0$ almost surely $\implies \lim_{n\to\infty}\mathbb{E}\left[\frac{|X_n|}{1+|X_n|}\right]=0$

My thoughts: $\lim_{n\to\infty}X_n=0 \implies \lim_{n\to\infty}1+|X_n|=1 \implies \lim_{n\to\infty}\frac{1}{1+|X_n|}=1$ (I think?) but then I'm not sure where to go from here.

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Use the dominated convergence theorem. The argument inside the expectation is bounded by $1$, which is integrable. So you may take the limit inside of the expectation, so it would suffice to show that the limit of the argument is $0$.

You are correct in stating that $|X_n| \to 0$ a.s. also. If your probability space is $(\Omega, \mathscr{F}, \mathbb{P})$, then take any $\omega \in \Omega$ for which $X_n(\omega) \to0$ as $n \to \infty$ to convince yourself. However, by the algebra of limits: \begin{equation} \lim_{n \to \infty} \dfrac{|X_n(\omega)|}{1+|X_n(\omega)|} = \lim_{n \to \infty} |X_n(\omega)| \times \lim_{n \to \infty} \dfrac{1}{1+|X_n(\omega)|} = 0. \end{equation}

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Observe that $\frac{|X_n|}{1+|X_n|} \leq 1$, then $E \left[\frac{|X_n|}{1+|X_n|}\right] \leq E[1]=1$ for any $n \in \mathbb{N}$. By dominated convergence you got $$E\left [\frac{|X_n|}{1+|X_n|}\right] \xrightarrow{n} 0 $$

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