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If ${X_n}$ is a sequence of random variables defined on the same probability space, does $\lim_{n\to\infty}\mathbb{E}(|X_n|)=0$ imply that $X_n$ converges to $0$ almost surely? I have a sense that this is true -- since the expected value of the absolute value of the sequence (which is all positive) converges to 0, the sequence itself converges to 0, intuitively -- but recalling the formal definition of almost sure convergence ($X_n\to X$ iff $\mathbb{P}\left(\{\omega\in\Omega|\lim_n X_n(\omega)=X(\omega)\})=1\right)$.

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Hint: Counterexample

Consider the probability space $((0,1), \mathcal{B},Leb)$and the indicator function

$$1_{(a,b)}(\omega) = \begin{cases}1, \,\, \omega \in(a,b)\\0, \,\,\omega \notin (a,b) \end{cases} $$

Let

$$X_1 = 1_{(0,1/2)}\, , \,\,X_2 = 1_{(1/2,1)} \,, \,\, X_3 = 1_{(0,1/4)} \, , \,\, X_4 = 1_{(1/4,1/2)}\, , \, \, X_5 = 1_{(1/2,3/4)} \, , X_6 = 1_{(3/4,1)} \, , \ldots $$

Now show $E(|X_k|) \to 0$ and for almost every $\omega \in (0,1)$ and for every $n \in {N}$ there exists $n_1, n_2 > n$ such that $X_{n_1}(\omega) = 1$ and $X_{n_2}(\omega) = 0$.

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  • $\begingroup$ To be strictly accurate, note that your definition gives $X_n(\omega)=0$ at $\omega=1/2$ for all $n\geq 1$. $\endgroup$
    – user940
    Commented Nov 23, 2016 at 21:34
  • $\begingroup$ @Byron Schmuland: Thanks. I added the words almost every $\omega$ to the hint. $\endgroup$
    – RRL
    Commented Nov 23, 2016 at 21:40

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