Show $P\left[AI have a question about exponential distribution and conditional probability.
Let $(\Omega, \mathcal{F}, P)$ be a probability space and $\mathcal{G}$ be a sub $\sigma$ algebra of $\mathcal{F}$. Let $Z$ be a non-negative random variable on $(\Omega,\mathcal{F},P)$ which is of the exponential distribution with mean $1$ and $A$ be a non-negative random variable on $(\Omega, \mathcal{G},P)$. We may assume $Z,\mathcal{G}$ are independent if necessary.
Question
I want to show $P\left[A<Z \left| \right. \mathcal{G} \right]=e^{-A}$. 
My idea
If $A=c$ (constant), $P \left[c<Z \right]=e^{-c}$. But I don't know how to deal with $\mathcal{G}$. 
If you know, please tell me.
Thank you in advance.

WARNING: The following argument attempts to make use of conditional expectations which I do not understand entirely rigorously (as @Did can confirm).
$P\left[A<Z \left| \right. \mathcal{G} \right]=E\left[1_{\{A<Z\}} \left| \right. \mathcal{G} \right]$ just by definition.
Moreover, a regular conditional distribution such that the following makes sense should exist because we are working on the real line.
(Specifically I believe Theorem 1 applies.)
Let $B \in \mathcal{G}$. Then:
$$\int_B e^{-A} = \int_B \int_{R_{\ge0}} e^{-c}\mathrm{d}P(A=c)=\int_B \int_{R_{\ge0}} P[c<Z]\mathrm{d}P(A=c) = \int_B P[A<Z] = \int_B P[A<Z|\mathcal{G}]$$
Thus, by @Did's comment above:
$$e^{-A}=P[A<Z|\mathcal{G}]\ \text{a.s.}$$
