Concerning the proof of the martingale convergence theorem 7.1.4 on page 341 of Ethier and Kurtz (1986), at the beginning of the proof for case (a), it is written that:
"Since (1.19) implies $\eta_n\to\infty$ in probability, the convergence of $M_n$ is equivalent to the convergence of $\tilde M_n = M_n(\cdot\wedge\eta_n)$."
where $\eta_n$ is a sequence of stopping time, and $M_n$ is a sequence of martingales in $(D[0,\infty))^d$, and the "convergence" is the weak convergence. I can see why (1.19) implies $\eta_n\to\infty$, but I don't understand why "the convergence of $M_n$ is equivalent to the convergence of $\tilde M_n = M_n(\cdot\wedge\eta_n)$." It seems that this should be obvious because they haven't added any reference. The only relevant thing I have found in their book so far is Thm 4.6.3 on page 219, where the uniqueness of the solution of martingale problems is constructed by the uniqueness of the solution of the stopped martingale problems, but this theorem does not concern the weak convergence.
Does anybody have any idea? Thank you for your help.
The reason of ``the convergence of $M_n$ is equivalent to the convergence of $\tilde{M}_n=M_n(\cdot\wedge\eta_n)$'' is the following fact: If $\text{pr-}\lim_{n\to\infty}d(M_n,\tilde{M}_n)=0$, then the convergence of $M_n$ is equivalence of the convergence of $\tilde{M}_n$, where $d(x,y)$ is the distance in $D_E([0,\infty)$ defined in p.117(5.2). (cf. P. Billingsley, Convergence of Probability Measures, 2ed. Th3.1 p.27). Now for the $d(M_n,\tilde{M}_n)$ we have following estimation $$ d(M_n,\tilde{M}_n)\le \int_0^\infty e^{-u}\sup_{0\le t\le u}[r(M_n(t),M_n^{\eta_n}(t))\wedge 1]du \le e^{-\eta_n} \overset{\mathsf{P}}\to 0.\quad\text{as } n\to\infty.$$