Likelihood function for Type I censoring

Question

I'm reading the textbook Survival Analysis: Techniques for Censored and Truncated Data by Klein and Moeschberger, and in Chapter 3.5 it says

Data from experiments involving right censoring can be conveniently represented by pairs of random variables $(T, C_r)$, where $C_r$ indicates whether the lifetime X is observed ($\delta = 1$) or not ($\delta = 0$), and T is equal to X if the lifetime is observed and to $C_r$ if it is right-censored, i.e., $T = \min(X, C_r)$. Details of constructing the likelihood function for Type I censoring are as follows. For $\delta = 0$, it can be seen that $$Pr(T, \delta = 0) = Pr(T = C_r|\delta = 0) Pr(\delta = 0) = Pr(\delta = 0) = Pr(X > C_r) = S(C_r)$$ Also, for $\delta = 1$, $\begin{aligned} Pr(T,\delta = 1) &= Pr(T=X|\delta = 1)Pr(\delta = 1) \\ &= Pr (X = T | X \leq C_r) Pr (X \leq Cr) \\ &= \left(\frac{f(t)}{1-S(Cr)}\right) \left(1-S(Cr)\right) \\ &= f(t) \end{aligned}$

I'm confused about the following:

1. when $\delta = 0$, from step 2 to step 3, it seems to let $Pr(T = C_r|\delta = 0) = 1$, but I'm not sure why is that

2. when $\delta = 1$, we have $Pr (X = T | X \leq C_r) = \frac{Pr(X = T = t \land X \leq C_r)}{Pr( X \leq C_r)}$, but why is the numerator equals to $f(t)$?

Answer 1

$\bullet$ Conditionally when $\delta = 0,~ T= C_r~\textrm{a.s.}$ and hence $\mathbb P[T = C_r|\delta = 0] = 1. $

$\bullet$ Since, conditionally $\delta = 1\implies X\leq C_r$ and $T = t \leq C_r$ by definition, then $\mathbb P[(X = T = t) \land (X \leq C_r)] = f(t). $