How does one read the $1_{[T_n,T_{n+1})} (t)$ in exploding random process? How does one read the $1_{[T_n,T_{n+1})} (t)$ in exploding random process?
I.e. in $X_t = \sum_{n=1}^{\infty} n 1_{[T_n,T_{n+1})} (t)$.
I read that $t$ should belong to $[T_n,T_{n+1})$ for it to take value $1$.
But since $T_n=S_1+...+S_n$, $S_i \sim Exp(\lambda_i)$.
Then I don't understand what it means for some $t$ to belong "between sums of random exponential variables".
 A: If it helps, think of $t$ as a fixed number for now (say, $t=4$). It is not a random quantity.
$1_{[T_n, T_{n+1})}(t)$ is a random variable that depends on the random variables $T_n$ and $T_{n+1}$. It equals $1$ when $T_n \le t < T_{n+1}$, and equals zero otherwise. So in the case $t=4$, this random variable equals $1$ when the event $T_n \le 4 < T_{n+1}$ holds, and equal zero otherwise.
Then the whole sum $\sum_{n\ge 1} n 1_{[T_n, T_{n+1})}(t)$ is a sum of random variables. Different choices of $t$ will give different random variables.

In your particular case, you can think of $T_n$ as the arrival time of the $n$th customer to your store (and $S_i$ as the time between the arrival times of the $(i-1)$st and $i$th customers). Then $1_{[T_n, T_{n+1})}(t)$ equals $1$ if the $n$th customer arrives before or at time $t$ and if the $(n+1)$st customer arrives after time $t$. All terms in the sum $\sum_{n \ge 1} n 1_{[T_n, T_{n +1})}(t)$ will equal zero except for one (since for a fixed $t$, there is only one $n$ such that the event $\{T_n \le t < T_{n+1}\}$ occurs). In particular, $X_t$ will be the number of customers before/at time $t$.
