Following my previous question, let $0 \lt c \lt 1$ and consider the statements $$\mathbb{P}[\{\omega\in \Omega: \forall \epsilon\gt 0 \ \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = c, \tag{1}$$ $$\forall \epsilon\gt 0 ,\ \ \ \ \mathbb{P}[\{\omega\in \Omega: \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}] = c. \tag{2}$$ I think in general the equivalence $(1) \iff (2)$ doesn't hold since the proof for the case $c = 1$ heavily relies on the value of probability and the proof doesn't seem to work for any other value except $c = 1$ (and maybe $c = 0$) but I couldn't construct a counterexample. Is my guess correct? If the answer is yes, then what counterexample can be given?
Edit: Andrew's answer shows that $(2) \implies (1)$ is true. Let $A_\epsilon = \{\omega\in \Omega: \exists N\in \mathbb{N} \ \forall n\ge N \ |X_n(\omega) - X(\omega)| \lt \epsilon \}$ and $A = \cap_{\epsilon >0} A_{\epsilon}$. We want to show that $$\forall \epsilon \gt 0, \ \ P(A_\epsilon) = c \implies P(A) = c$$ holds. Define the conditional probability $\tilde{P}(\cdot) = P(\cdot \cap A_1)/P(A_1)$. Note that $\tilde{P}(\cdot)$ is well-defined since we have $P(A_1) \gt 0$: $$\forall \epsilon \gt 0, \ \ P(A_\epsilon) = c \implies P(A_1) = c \gt 0$$ Note that $(A_\epsilon)_{\epsilon \gt 0}$ is a nondecreasing family of subsets which means that $A_{\epsilon_1} \subseteq A_{\epsilon_2}$ for every $\epsilon_1 \le \epsilon_2$. So for $\epsilon \ge 1$ we have $A_1 \cap A_\epsilon = A_1$ and for $\epsilon \lt 1$ we have $A_1 \cap A_\epsilon = A_\epsilon$. This shows that $$\tilde{P}(A_{\epsilon}) = \begin{cases} P(A_1)/P(A_1) = 1, & \epsilon \ge 1 \\ P(A_\epsilon)/P(A_1) = c/c = 1, & \epsilon \lt 1 \end{cases}$$ By the previous result, this shows that $\tilde{P}(A) = 1$ and so $P(A) = P(A \cap A_1) = P(A_1) = c$. What about the other direction? Does it hold that $(1) \implies (2)$?
We use the notation in the comments. Define the probability measure $\tilde{\mathbb P}(\cdot) = \mathbb P(\cdot \cap A_1)/\mathbb P(A_1)$. Note $\tilde{\mathbb P}(A_\epsilon) = 1$ for all $\epsilon>0$. If $\epsilon\geq 1$, then $A_\epsilon\cap A_1 = A_1$. If $\epsilon \leq 1$, then per our hypothesis $$\tilde{\mathbb P}(A_\epsilon) = \mathbb P(A_\epsilon \cap A_1)/\mathbb P(A_1) = \mathbb P(A_\epsilon)/\mathbb P(A_1) = c/c=1$$ Continuity of measure therefore implies $X_n \to X$ $\tilde{\mathbb P}$ -almost surely. We conclude $\mathbb P(\lim X_n = X) = c$.