# Probability for all epsilon is one implies for all epsilon probability is one

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.

• What do you mean by this notation?
– Tim
Commented Mar 28, 2023 at 20:14
• 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 "$\forall \epsilon\gt0,A_\epsilon$"?
– whuber
Commented Mar 28, 2023 at 20:29
• 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. Commented Mar 28, 2023 at 20:45
• Technically, you should answer this question first: given $A_\epsilon \in \mathscr{F}$ for all $\epsilon > 0$, can you guarantee that $\cap_{\epsilon > 0}A_\epsilon \in \mathscr{F}$? Commented Mar 28, 2023 at 20:55
• 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?
– whuber
Commented Mar 28, 2023 at 21:02

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