I am struggling to work out if the following is true.

Let $\{A_\epsilon\}$ be an indexed set of events in a probability space. Does it hold that:

$$\forall \epsilon > 0, \mathbb{P}[A_\epsilon]=1 \implies \mathbb{P}\bigg[ \bigcap_{\epsilon > 0} A_\epsilon\bigg]=1$$

If not, can I add any conditions (such as the $A_\epsilon$ are decreasing or increasing) to guarantee the result?

Edit: The origin of this question is from an almost sure convergence proof. The proof showed that:

$$\forall \epsilon > 0, \mathbb{P}[\exists N \in \mathbb{N} \text{ such that } |X_n-X|\leq \epsilon]=1$$

but then the proof just jumped to:

$$\mathbb{P}[\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } |X_n-X|\leq \epsilon]=1$$

and that confused me a lot.

  • 2
    $\begingroup$ What do you mean by this notation? $\endgroup$
    – Tim
    Commented Mar 28, 2023 at 20:14
  • 1
    $\begingroup$ I, too, cannot make any sense of the right hand side. I am expecting the probability to be of an event, but what event is "$\forall \epsilon\gt0,A_\epsilon$"? $\endgroup$
    – whuber
    Commented Mar 28, 2023 at 20:29
  • $\begingroup$ Sorry for the confusion, I was thinking in term of logic events "for all epsilon, $A_\epsilon$ happens" - I realise this was bad notation and I've changed the notation. $\endgroup$
    – MasonTep
    Commented Mar 28, 2023 at 20:45
  • $\begingroup$ Technically, you should answer this question first: given $A_\epsilon \in \mathscr{F}$ for all $\epsilon > 0$, can you guarantee that $\cap_{\epsilon > 0}A_\epsilon \in \mathscr{F}$? $\endgroup$
    – Zhanxiong
    Commented Mar 28, 2023 at 20:55
  • $\begingroup$ Just to make sure I'm understanding: suppose the probability space is the real interval $(0,1)$ with the uniform measure and that the indexing set for $\epsilon$ is also $(0,1),$ where $A_\epsilon = (0,1)\setminus\{\epsilon\}$ (the interval with the number $\epsilon$ removed). Notice that $\Pr(A_\epsilon)=1$ for all $\epsilon$ but $\cap A_\epsilon$ is empty. Would this be a valid example of what you are asking about? $\endgroup$
    – whuber
    Commented Mar 28, 2023 at 21:02

1 Answer 1


The statement from the proof in the edited part of the question has $A_{\epsilon_1}\subset A_{\epsilon_2}$ for $\epsilon_1<\epsilon_2$. This means that the intersection can be written as countable intersection for, say, $\epsilon\in\{\frac{1}{n}\ :\ n\in\mathbb{N}\}$. For countable intersections, the statement holds, see Countable intersection of almost sure events is also almost sure

The result will generally hold for increasing or decreasing sequences, as here the countable intersection is the same as the uncountable one.

  • $\begingroup$ I see that this has been already almost completely done in the comments to the question. I have only realised this after writing the answer, but of course it makes sense to have this as an answer anyway. $\endgroup$ Commented Mar 28, 2023 at 23:53
  • $\begingroup$ No this is really helpful! Thank you @Christian Hennig! This was what I was looking for! $\endgroup$
    – MasonTep
    Commented Mar 29, 2023 at 13:49

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