I cannot understand why the following expression is true
$\operatorname{Pr}(A \in D, B \leq t)=\int_{0}^{t} f_{B}(z) \operatorname{Pr}(A \in D \mid B=z) \mathrm{d} z$ (I)
From my limited understanding, using rules of probability:
$\operatorname{Pr}(A \in D, B \leq t)= \operatorname{Pr}(A \in D| B \leq t) * \operatorname{Pr}(B \leq t) = \operatorname{Pr}(A \in D| B \leq t) * \int_{0}^{t} f_{B}(z) \mathrm{d} z$ (II)
where $f_{B}(z)$ is the pdf.
Is (II) correct? If yes, then how do I go from (II) to (I)? If not, then why is (I) correct? Is there a derivation?
Comparing both equations, feels like in (I), the conditional term is somehow pulled in the integration and the $\leq t$ is changed to =. Can someone explain why (I) is correct mathematically?
I need to get to (I) to understand a proof.
This is also on pg. 175 of Durek Probability in the expansion of $P(N_t = n)$.
Thanks in advance!