Let $(\Omega \times \Omega ', \mathcal F\times \mathcal A,\mathbb P)$ a probability space. In fact $\mathcal A=\sigma \{\alpha_i\mid i\in\mathbb N\}$ where $\alpha _i\in [0,1]$ are random variable (iid but it's not important for my question). Let $$X:\Omega\times \Omega '\longrightarrow \mathbb R$$ a random variable. What does $\mathbb P\{X\leq x\mid \mathcal A\}$ mean ? So, I know that it's a r.v. that is $\mathcal A-$ measurable and it's defined by $$\mathbb E[\boldsymbol 1_{\{X\geq x\}}\mid \mathcal A].$$
Q1) Now, since $\mathcal A$ has only element in $[0,1]$ does it imply that $\mathbb P\{X\leq x\mid \mathcal A\}\in [0,1]$ a.s. ?
Q2) If for example $\mathcal A$ was the $\sigma -$algebra of borel set, could we have $\mathbb P\{X\geq x\mid \mathcal A\}$ ? (I know it look stupide for a probability to be bigger than $1$, but since it's a r.v., after all, why not...)
Q3) If I have that $\mathbb P\{X\leq x\mid \mathcal A\}=f(\alpha _1,...,\alpha _n)$ for example, in this situation (since it's $\mathcal A-$ measurable) can the $\alpha _i$ be seen as "fix number" ? I know they are not, but in proof, I often see : fix a sequence "$\alpha _0,\alpha _1,...$", so maybe in this situation they can be seen as fixed number (in the context conditional by $\mathcal A$)