Show $P\left[A<Z \mid \mathcal{G} \right]=e^{-A}$ for $Z$ standard exponential and $A$ nonnegative $\mathcal G$-measurable

Question

I 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.

Answer 1

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:

$$e^{-A}=P[A<Z|\mathcal{G}]\ \text{a.s.}$$