I'm learning about continuous time markov chains, in the text I am reading, they are setting up the discussion by talking about a series of alarm clocks that are set:
Suppose $T_1,...,T_n$ are independent random variables, each exponential with rates $b_1,...,b_n$, respectively. Intuitively, we can think of $n$ alarm clocks which will go off at the times $T_1,...,T_n$. Consider the first time when any of the alarm clocks goes off; more precisely consider the random variable $$T = min\{T_1,...,T_n\}$$ Note that $$\mathbb{P}\{T \ge t\} = \mathbb{P}\{T_1 \ge t, ..., T_n \ge t\} = \mathbb{P}\{T_1 \ge t\} \mathbb{P}\{T_2 \ge t\}...\mathbb{P}\{T_n \ge t\} = e^{-b_1t}e^{-b_2t}...e^{-b_nt} = e^{-(b_1+b_2+...+b_n)t}$$
In other words, $T$ has an exponential distribution with parameter $b_1+...+b_n$. Moreover, it is easy to give the probabilities for which of the clocks goes off first, $$\mathbb{P}\{T = t\} = \int_0^\infty\mathbb{P}\{T_2 > t\,...,T_n > t\} d\mathbb{P}\{T_1 = t\} = \int_0^\infty e^{-(b_2+...+b_n)t}b_1 e^{-b_1t}dt = \frac{b_1}{b_1+...+b_n}$$
Now, I'm comfortable with everything except the expression $d\mathbb{P}\{T_1 = t\}$, and how you are extracting $dt$ after you write out the expressions for the probability functions (e.g. $\mathbb{P}\{T_1 = t\} = b_1e^{-b_1t})$
I realized this is probably a fundamental lack of knowledge of calculus on my part, but any help would be greatly appreciated.
A nonnegative random variable $X$ is exponentially distributed with parameter $\lambda$ if $$P[X\geq t]=e^{-\lambda t}\qquad(t\geq0)\ .$$ It has a probability density $$f_X(t):=\lim_{h\to0+}{P[t\leq X\leq t+h]\over h}=\lim_{h\to0+}{e^{-\lambda t}-e^{-\lambda(t+h)}\over h}=\lambda\>e^{-\lambda t}\ .$$ Your last displayed formula then should read $$\eqalign{P[T=T_1]&=\int_0^\infty P[T_2>t, T_3>t,\ldots,T_n>T]\>f_{T_1}(t)\>dt\cr &=\int_0^\infty e^{-(b_2+\ldots+b_n)t}\> b_1e^{-b_1t}\>dt={b_1\over b_1+b_2+\ldots+b_n}\ .\cr}$$ This means that your cryptic $dP[T_1=t]$ expands without much ado to $f_{T_1}(t)\>dt$.