I've seen these two definitions of almost sure convergence:

  • $\mathbb{P}\left(\lim _{n \rightarrow \infty} X_n=X\right)=1$
  • The sequence $X_n$ converges almost surely to $X$ if there exists a sequence of random variables $\Delta_n$ such that $d\left(X_n, X\right) \leq \Delta_n$ and $\Delta_n \stackrel{\text { as }}{\rightarrow} 0$.

The first of these came from Wikipedia and the second came from Asymptotic Statistics by A.W. Van der Vaart. Are these equivalent? I'm a little confused.

  • 1
    $\begingroup$ If you read the text carefully, the second definition is the definition of $X_n \overset{as*}{\to} X$ with respect to the outer probability measure $P^*$, not the standard definition of converge almost surely (with respect to the original probability measure $P$). On the other hand, if $X_n$ and $X$ are all real-valued random variables, then these two definitions are indeed equivalent. $\endgroup$
    – Zhanxiong
    Mar 30 at 2:32
  • $\begingroup$ @Zhanxiong, if you wish, you can make it as an answer, for I think it directly hits at the poster's confusion. $\endgroup$ Mar 30 at 2:40
  • 1
    $\begingroup$ @User1865345 Followed your request, my friend. $\endgroup$
    – Zhanxiong
    Mar 30 at 3:34

1 Answer 1


As I commented under the question, the second definition looks more contrived because it wants to accommodate the situation that $X_n$ and $X$ are not measurable or $d(X_n, X)$ is not measurable (for more details, see Section 18.2 in Asymptotic Statistics). In this situation, the set $[\lim_{n \to \infty}X_n = X]$ may not be an event so that the notation "$P[\lim_{n \to \infty}X_n = X]$" may become meaningless.

That said, the two definitions are indeed equivalent when $X_n$, $X$, and $d(X_n, X)$ are all measurable. For example, when $X_n$ and $X$ are real-valued random variables and $d(x, y) = |x - y|$. This case can be proved as follows.

In this thread, it was shown that $Y_n$ converges to $Y$ almost surely if and only if for any $\epsilon > 0$, it holds that \begin{align} P[\cap_{k \geq 1}\cup_{n \geq k}[|Y_n - Y| \geq \epsilon]] = 0. \tag{1} \end{align}

Suppose $|X_n - X| \leq \Delta_n$ and $\Delta_n$ converges to $0$ almost surely, it then follows by $(1)$ that (set $Y_n = \Delta_n$, $Y = 0$) for any $\epsilon > 0$:
\begin{align} P[\cap_{k \geq 1}\cup_{n \geq k}[|X_n - X| \geq \epsilon]] \leq & P[\cap_{k \geq 1}\cup_{n \geq k}[\Delta_n \geq \epsilon]] = 0, \end{align} which implies (by the other direction of $(1)$) that $X_n$ converges to $X$ almost surely. This shows that the second definition implies the first definition.

Conversely, suppose $X_n$ converges to $X$ almost surely, then simply taking $\Delta_n = |X_n - X|$ (or if you don't want to be that extreme, take $\Delta_n = 2|X_n - X|$, say) meets all the requirements in the second definition. This shows the first definition implies the second definition. In conclusion, these two definitions are equivalent.

  • $\begingroup$ Thank you @Zhanxoing! That helps a lot! It's really appreciated! $\endgroup$
    – JDoe2
    Mar 30 at 11:59

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