How to prove $P(T_n\le t< T_{n+1}) = \int_0^t P(T_n = s) P(\xi_{n+1} > t-s) ds$? Here is a proof from Rick Durrett's Probability, Poisson process, and I am confused with it.
Here $\xi_1,\xi_2\cdots$ are independent random varibales with $P(\xi_i > t) = e^{-\lambda t}$.
$T_n = \sum_{k=1}^n \xi_k$, $N_t=\sup\{n: T_n \le t\}$. Now we have $P(N_t = 0) = P(T_1>t)=e^{-\lambda t}$ and for $n\ge 1$,
$P(N_t=n)=P(T_n\le t< T_{n+1}) = \int_0^t P(T_n = s) P(\xi_{n+1} > t-s) ds$ ...
I am confused about the second equality, how to get it? How to get a integral form from the probability?
 A: Note the following, you are looking for the probability of the event $\{N_t = n\} = \{T_n \leq t < T_{n+1}\}$. Hence, for any $s \leq t$, if $T_n = s$, and $\xi_{n+1}>t-s$, this means that $T_{n+1} = T_n + \xi_{n+1} > s+(t-s)=t$.
Therefore, I hope it is now clear that $\{T_n = s, \xi_{n+1} > t-s\} \subset \{T_n \leq t < T_n+1\}$. And
$$\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\}  = \{T_n \leq t < T_n+1\}$$
Since the events are the same, we know that
$$P(\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} ) = P(\{T_n \leq t < T_n+1\})$$
Finally, since $\xi_{n+1}$ is independent of $T_n$, we have:
$$
P(\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} )= \int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds
$$
To prove the identity above, we assume that $P\circ T_n^{-1}$ is absolutely continuous with respect to the Lebesgue measure. Hence,
$$
\int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds =
\int_{T_n^{-1}([0,t])}P(\xi_{n+1}>t-s)d(P\circ T_n^{-1})(s)
$$
Now, we can partition $[0,t]$ in $k \in \mathbb N$ subdivisions (i.e. $[0,t/k)\cup...\cup[t(k-1)/k,t]$).
As $k \to \infty$, the set
$$A_k:= 
\bigcup_{i\in \{1,...,k\}}
\{T_n \in \big[t\cdot (i-1)/k,t\cdot i/k \big ),
\quad \xi_{n+1} > t-(i\cdot t/k)\}
\\
A_k
\to
\bigcup_{s\in[0,t]}\{T_n = s, \xi_{n+1} > t-s\} = A
$$
By the continuity of probability, then
$P(A_k) \to P(A)$.
The only thing left to do is to prove that
$$P(A_k) \to
\int^t_0 P(T_n = s)P(\xi_{n+1}> t-s)ds 
$$
Let's call $I_i = [t\cdot (i-1)/k, t \cdot i/k)$, and define a simple
function
$$f_k = \sum_{i=1}^k P(\xi_{n+1}> t-t\cdot i/k) \mathbb I_{I_i}$$
Hence, we have:
$$
P(A_k) = \sum^k_{i=1}P(T_n \in I_i)P(\xi_{n+1}> t-t\cdot i/k)=
\int_{T_n^{-1}([0,t])}f_kd(P\circ T_n^{-1})(s)
$$
Finally, since $f_k(s) \to f(s) = P(\xi_{n+1}>t-s)$. We can use the Dominated Convergence Theorem to conclude that:
$$
P(A_k) = \int_{T_n^{-1}([0,t])}f_kd(P\circ T_n^{-1})(s)
\to
\int_{T_n^{-1}([0,t])}P(\xi_{n+1}>t-s)d(P\circ T_n^{-1})(s)
$$
