Survivor function of a variable that has discrete and continuous components I'm currently reading The Statistical Analysis of Failure Time Data by Kalbfleisch and Prentice and had trouble at arriving at the expression for the survivor function of a random variable $T$ having both discrete and continuous components. The setup is the following:
Let $T$ be a random variable on $[0,\infty)$ with survivor function $F(t)=P(T>t)$. Then 


*

*if $T$ is absolutely continuous with density $f$, then the hazard function $\lambda$ can be defined as 
$$
\lambda(t):=\lim_{h\to 0^+}\frac{P(t\leq T<t+h\mid T\geq t)}{h}=\frac{f(t)}{F(t)}
$$ for $t\geq 0$, and hence we have
$$
F(t)=\exp\left(-\int_0^t\lambda(s)\,\mathrm ds\right),\quad t\geq 0.
$$

*if $T$ is discrete taking on the values $0\leq a_1<a_2<\cdots$, then we define the hazard at $a_i$ as
$$
\lambda_i=P(T=a_i\mid T\geq a_i),\quad i=1,2,\ldots.
$$
Then we can show that
$$
F(t)=\prod_{j\mid a_j\leq t}(1-\lambda_j),\quad t\geq 0.
$$
These expressions for the survivor functions I am ok with. Now they write the following:

More generally, the distribution of $T$ may have both discrete and continuous components. In this case, the hazard function can be defined to have the continuous component $\lambda_c(t)$ and discrete components $\lambda_1,\lambda_2,\ldots$ at the discrete times $a_1<a_2<\cdots$. The overall survivor function can then be written
  $$
F(t)=\exp\left(-\int_0^t\lambda_c(s)\,\mathrm ds\right)\prod_{j\mid a_j\leq t}(1-\lambda_j).\tag{1}
$$

That $T$ has both discrete and continuous components means that the distribution of $T$ is of the form
$$
P_T(\mathrm dx)=f_c(x) \lambda(\mathrm dx)+\sum_{j=1}^\infty b_j \delta_{a_j}(\mathrm dx)
$$
or equivalently
$$
P(T\in A)=\int_A f_c(x)\,\mathrm dx+\sum_{j\mid a_j\in A} b_j
$$
for some sequence $a_1<a_2<\cdots$ and $b_i\in (0,1)$ and some non-negative measurable function $f_c$ with $\int_0^\infty f_c\,\mathrm d\lambda+\sum_{i=1}^\infty b_j=1$. If we define
$$
\lambda_c(t)=\frac{f_c(t)}{P(T\geq t)}=\frac{f_c(t)}{F(t)},\quad t\neq a_i,
$$
and
$$
\lambda_i=P(T=a_i\mid T\geq a_i),
$$
then how do I show (and is it even true) that the survivor function of $T$ is given by $(1)$?
 A: The function $F(t) = \mathsf P(T>t) = 1-\mathsf P(T\leq t)$ is clearly of RCLL class on $[0,\infty)$. As a result, the definitions of continuous part of the hazard function $\lambda_c$ and discrete parts allow you computing $F$ by integrating $\lambda_c$ in between of the jumps, and applying jump conditions at $t = a_j$. The latter have the following shape:
$$
  \lambda_j = \mathsf P(T = a_j\mid T\geq a_j) = \frac{F(a_j-) - F(a_j)}{F(a_j-)}\implies F(a_j) = F(a_j-)(1-\lambda_j)
$$
where $F(t-):=\lim_{s\uparrow t}F(s)$.
A: This may be very late but I would like to give my two cents on this question.
Suppose $\mu$ is a probability measure on $((0,\infty),\mathscr{B}((0,\infty))$ and let $F(x):=\mu(0,x]$. The Integrated Hazard Function $Q$ of $\mu$ is defined as
$$
Q(t)=\int_{(0,t]}\frac{1}{1-F(x-)}\mu(dx).
$$
The function $S(t):=1-F(t)$ is a right--continuous monotone nonincreasing function. $Q$ is a right--continuous monotone nondecreasing function whose associated (Lebesgue-Stieltjes) measure $\mu_Q\ll\mu$ satisfies
$$
\begin{align}
\mu_{Q}(\{x\})&=\Delta Q(x)=\frac{\Delta F(x)}{S(x-)}\\
\mu_{Q_c}(dx)&=\frac{1}{S(x-)}\mu_{F_c}(dx)\\
S(x-)\mu_Q(dx)&=\mu(dx)=\mu_F(dx).
\end{align}
$$
where $F_c$ and $Q_c$ is the continuous part of $F$ and $Q$ respectively. Then,
$Q$ and $F$ have the same points of discontinuity $\{x_j:j\in I\}$, and since $S(t)=1-F(t)=1-\int_{(0,t]}\mu(dx)$,
$$
\begin{align}
S(t)=S(0)-\int_{(0,t]}S(x-)\mu_Q(dx)\tag{1}\label{one}
\end{align}
$$
We will show that $S$ is the unique solution to $\eqref{one}$ that is bounded in any bounded set, and that
$$
S(t)=\exp\big(-Q_c(t)\big)\prod_{0<x_j\leq t}
(1-\Delta Q(x_j)).
$$

The proof of this will be a consequence of the following theorem:
Theorem: Let $F$ be a right--continuous  monotone nondecreasing function in $[0,\infty)$ and let $\mu_F$ be the unique measure on $(0,\infty)$ such that $\mu\big((a,b]\big)=F(b)-F(a)$. Let $\{x_j:j\in\mathbb{N}\}$ be the sequence of all discontinuities of $F$. If $v\in\mathcal{L}^{loc}_1(\mu_F)$ then,  for any  number $H_0\geq0$ the function
$$
\begin{align}
H(t)=H_0\exp\Big(\int_{(0,t]}v(x)\mu_{F_c}(dx)\Big)\prod_{0<x_j\leq t}(1+v(x_j)\Delta F(x_j))\tag{2}\label{expo-form}
\end{align}
$$
is the unique solution in $t\geq0$ of the integral equation
\begin{align}
\label{integro-exp}
H(t)=H(0)+\int_{(0,t]}H(x-)v(x)\mu_F(dx)
\end{align}
satisfying $\|H\mathbb{1}_{(0,t]}\|_u<\infty$ for all $t>0$.

The formula quoted in the OP is $\eqref{one}$, and existence and uniqueness are obtained by the Theorem above with $v\equiv-1$.

Since formula $\eqref{expo-form}$ appears often in applications Survival analysis and reliability theory, I think I is worthwhile to present a proof. This will be based entirely on Lebesgue integration by parts.
Preliminary notation:
For any real valued function $F$ on an interval $I$, denote by $\mu_F$ the Lebesgue-Stieltjes measure generated by $F$, so $\mu_F\big((a,b]\big)=F(b)-F(a)$ for all $[a,b]\subset I$.

*

*Recall that for any real valued functions $F$, $G$ of local finite variation in some interval $I$
$$
\int_{(a,b]} F(t)\,\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\,\mu_F(dt)
$$
for all $[a,b]\subset I$. This formula may be denoted as
$$
d(FG)=F\,dG+ G_-\,dF
$$
where $G_-(t):=G(t-)$ and $dF(x):=\mu_F(dx)$, that is $dF\big((a,b]\big)=F(b)-F(a)$.


*If $G$ is a continuous function of locally finite variation, then
$$ dG^n = n G^{n-1}(t)\,dG$$
This can be shown by induction. For $n=1$ is valid. For $n\geq1$
$$ d(G^{n+1})=d(G^n\,G)=G\,dG^n + G^n\,dG=nG^n\,dG+ G^n\,dG=(n+1) G^n\,dG$$
From this, we obtain the well known exponential formula for continuous measures:
$$\begin{align}
 d e^G(t) = e^{G(t)}\,dG(t):= e^{G(t)}\,d\mu_G(dt)\tag{3}\label{exp-for1}
\end{align}
$$
A technical result:
Lemma:
Suppose $G$ is right--continuous nondecreasing  in the interval $[0,T)$ $(0<T\leq\infty)$. Then, for any $n\in\mathbb{N}$
$$
\int_{(0,t]}G^{n-1}(s-)\mu_G(ds)\leq \frac{G^n(t)-G^n(0)}{n}\leq\int_{(0,t]}G^{n-1}(s)\mu_G(ds)
$$
for all $0<t<T$. (In differential notation,  $nG^{n-1}_-dG\leq dG^n\leq nG^{n-1}dG$.)
Here is a short proof:
For $n\in\mathbb{N}$, $G^n$ is right--continuous an nondecreasing and so, the associates Lebesgue--Stieltjes measure $\mu_{G^n}$ is nonnegative. Repeated application of integration by parts gives
$$
\begin{align}
  dG^n &= G^{n-1}_-\,dG + G\,dG^{n-1}=G^{n-1}_-\,dG + G (G^{n-2}_-\,dG + G\,dG^{n-2})\\
  &= (G^{n-1}_-+GG^{n-2}_- +\ldots + G^{n-1})\,dG
\end{align}
$$
in differential notation. As $G(s-)\leq G(s)$ for all $0<s\leq T$, we conclude that
$$
n G^{n-1}_-\,dG \leq dG^n\leq n G^{n-1}\,dG
$$
Proof of main Theorem:
As $v\in \mathcal{L}^{loc}_1(\mu_F)$, $v\in\mathcal{L}^{loc}_1(\mu_{F_I})$, and so
$$
\|v\mathbb{1}_{(0,t]}\|_{\mathcal{L}_1(\mu_{F_I})}=\sum_{0<x_j\leq t}|v(x_j)|\Delta F(x_j)<\infty.
$$
Consequently $H$ is bounded on each compact subinterval of $[0,\infty)$.
Let
$$
\begin{align}
G_1(t)&=H_0\prod_{0<x_j\leq t}(1+v(x_j)\Delta F(x_j))\\
G_2(t)&=\exp\Big(\int_{(0,t]}v(x)\mu_{F_c}(dx)\Big).
\end{align}
$$
$G_1$ is right--continuous pure jump function of bounded variation which changes  only at $x_j$; moreover,
$$
\begin{align}
\Delta G_1(x_j)=G(x_j)-G(x_j-)&=G(x_j-)\big(1+v(x_j)\Delta F(x_j)\big)-G(x_j-)\\
&= G(x_j-)v(x_j)\Delta F(x_j).
\end{align}
$$
$G_2$ is a continuous  monotone nondecreasing function and
$$
\begin{align}
\mu_{G_2}(dx)&=\exp\Big(\int_{(0,x]}v(y)\mu_{F_c}(dy)\Big)v(x)\mu_{F_c}(dx)\\
&= G_2(x)v(x)\mu_{F_c}(dx).
\end{align}
$$
Applying the integration by parts formula to $H(t)=G_1(t)G_2(t)$ gives
$$
\begin{align}
H(t)-H(0)&=\int_{(0,t]}G_1(x-)\mu_{G_2}(dx)+\int_{(0,t]}G_2(x)\mu_{G_1}(dx)\\
&= \int_{(0,t]}G_1(x-)G_2(x)v(x)\mu_{F_c}(dx)+
\sum_{0<x_j\leq t}G_2(x_j)G_1(x_j-)v(x_j)\Delta F(x_j)\\
&= \int_{(0,t]}H(x-)v(x)\mu_{F_c}(dx)+\int_{(0,t]}H(x-)v(x)\mu_{F_I}(dx)\\
&=\int_{(0,t]}H(x-)v(x)\mu_F(dx).
\end{align}
$$
It remains to prove uniqueness. Suppose $H_1$ and $H_2$ are two solutions and set $D=H_1-H_2$. Let $M:=\|D\mathbb{1}_{(0,t]}\|_u$ and $\Lambda(t)=\int_{(0,t]}|v(x)|\mu_F(dx)$. Then,
$$
|D(t)|\leq \int_{(0,t]}|D(x-)||v(x)|\mu_F(dx)\leq M\int_{(0,t]}|v(x)|\mu_{F}(dx)
= M\Lambda(t).
$$
As $\Lambda$ is nondecreasing and right continuous, $|D(x-)| \leq M\Lambda(x-)$. By the technical Lemma above
$$
\begin{align}
|D(t)|&\leq M\int_{(0,t]}\Lambda(x-) |v(x)|\mu_F(dx) = M\int_{(0,t]}\Lambda(x-)\mu_\Lambda(dx)\leq \frac{M}{2} \Lambda^2(t).
  \end{align}
$$
Continuing by induction we obtain
$|D(t)|\leq \frac{M}{n!}\Lambda^n(t)$. Letting $n\rightarrow\infty$ gives $|D(t)|=0$.
