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I have seen it from a book that almost surely convergence is equivalent to the following statement: for any $\epsilon>0$, there is an $n>0$, such that $P({|X_{j}-X|\geq\epsilon})=0$, for all $j\geq n.$ I doubt if it is right.

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Let $X_n: [0,1] \rightarrow \mathbb{R}$ be the random variable that is $1$ on $[0,1/n]$ and $0$ elsewhere. $\{X_i\}$ converges a.s. to $0$, but your condition does not hold. – ShawnD Sep 22 '11 at 6:53
up vote 4 down vote accepted


Almost sure convergence is $\Pr\!\left( \lim_{n\to\infty}\! X_n = X \right) = 1$.

These notes give the example

Let the sample space $S$ be the closed interval $[0, 1]$ with the uniform probability distribution. Define random variables $X_n(s) = s + s^n$ and $X(s) = s$. For every $s \in [0, 1)$, $s_n \rightarrow 0$ as $n \rightarrow \infty$ and $X_n(s) \rightarrow s = X(s)$. However, $X_n(1) = 2$ for every $n$ so $X_n(1)$ does not converge to $1 = X(1)$. But since the convergence occurs on the set $[0, 1)$ and $P([0, 1)) = 1$, $X_n$ converges to $X$ almost surely.

But for this example $|X_j-X|\geq\epsilon$ when $\epsilon^{1/j} \leq s \leq 1$, and that has positive probability $1-\epsilon^{1/j}$ for all positive $j$ and all $\epsilon \in (0,1)$. So this example has almost sure convergence but does not have the convergence you describe.

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nice counterexample, thank you – Jim Sep 22 '11 at 7:19

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