Let $X_1,X_2,X_3,\dots$ be a sequence of random variables, defined on the probability space $(\Omega, \mathcal{F} , \mathbb{P})$. We say that this sequence converges almost surely to random variable $X$ if, $$\mathbb{P}[\{\omega\in \Omega: \lim_{n\to \infty}X_n(\omega) = X(\omega) \}] = 1,$$ which is equivalent to $$\mathbb{P}[\{\omega\in \Omega: \forall \epsilon\gt 0 \ \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 1. \tag{1}$$ Sometimes the first quantifier is written at the beginning of sentence, i.e., $$\forall \epsilon\gt 0 ,\ \ \ \ \mathbb{P}[\{\omega\in \Omega: \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 1. \tag{2}$$ Are these definitions equivalent? In general, can we take the quantifiers out of $\mathbb{P}[.]$? I'm asking this question because in logic quantifiers should come before the remainder of the formula. The most formal definition which I've found (for the almost sure convergence) is as follows. For some $\mathcal{N}\in \mathcal{F}$ with $\mathbb{P}[\mathcal{N}] =0,$ $$\forall{\epsilon \gt 0} \ \forall \omega \not \in \mathcal{N} \ \exists N_{\epsilon, \omega} \ \forall n\ge N_{\epsilon, \omega} \ |X_n(\omega) - X(\omega)| \lt \epsilon. \tag{3}$$ In this definition $\mathbb{P}[\mathcal{N}] =0$ is before the quantifiers (except $\exists \mathcal{N} \in \mathcal{F}$ which occurs before $\mathbb{P}[\mathcal{N}] =0$), so I don't know how to fix it.
Edit: Actually, we can write two other definitions: $$\forall \epsilon\gt 0 \ \exists N\in \mathbb{N} ,\ \ \ \ \mathbb{P}[\{\omega\in \Omega:\forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 1 \\ \forall \epsilon\gt 0 \ \exists N\in \mathbb{N} \ \forall n\ge N,\ \ \ \ \mathbb{P}[\{\omega\in \Omega: |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 1$$When I translate these definitions to the words, it seems that they are all equivalent but I'm not sure. Also it would be interesting if we change the value of probability and ask whether the equivalence continues to hold. For example consider following propositions: $$\mathbb{P}[\{\omega\in \Omega: \forall \epsilon\gt 0 \ \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 0.45 \\ \forall \epsilon\gt 0 ,\ \ \ \ \mathbb{P}[\{\omega\in \Omega: \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = 0.45$$
The following proves $(1) \iff (2)$ and it follows the suggestions in the comments. To see this, note that we have $$\{X_n\to X\}=\bigcap_{k\in \mathbb{N}}\bigcup_{N\in \mathbb{N}}\bigcap_{n\geq N}\{|X_n-X|<1/k\}$$ From which it follows that for all $k \in \mathbb{N}$ $$\{X_n\to X\}\subseteq \bigcup_{N\in \mathbb{N}}\bigcap_{n\geq N}\{|X_n-X|<1/k\}=:A_k$$ And therefore $P(X_n\to X)=1$ implies $P(A_k)=1$ for all $k \in \mathbb{N}$. Now suppose $P(A_k)=1$ for all $k$. Define $$B_k:=A_1\cap (...)\cap A_k=\bigcap_{\ell \leq k}\bigcup_{N\in \mathbb{N}}\bigcap_{n\geq N}\{|X_n-X|<1/\ell\}$$ Then $P(B_k)=1$ for all $k \in \mathbb{N}$ because $$P(B_k^c)=P(A_1^c\cup (...)\cup A_k^c)\leq \sum_{\ell \leq k}P(A_\ell^c)=0$$ We also have that $B_k\supseteq B_{k+1}$ and $\cap_{k \in \mathbb{N}}B_k=\{X_n\to X\}$. But then by measure continuity $$P(X_n\to X)=\lim_{k\to \infty}P(B_k)=1$$