# I don't understand the proof of Corollary 4.8.7 in the book of Ethier and Kurtz

I'm trying to understand the proof of Corollary 8.7 of Chapter 4 in the book Markov Processes: Characterization and Convergence by Stewart N. Ethier, Thomas G. Kurtz. Here is the theorem and its proof:

And here are the relevant parts they refer to:

There are two things that I don't understand:

1. Why do they use $$\chi_{\left\{\:\tau_n\:>\:\color{red}{1}\:\right\}}$$ (indicator function of $$\left\{\tau_n>1\right\}$$) in the definition of $$\varphi_n$$. Wouldn't it make more sense to use $$\chi_{\left\{\:\tau_n\:>\:\color{red}{T}\:\right\}}$$ instead?
2. By definition, $$f_n(Y_n(t))-\int_0^tg_n(Y_n(s))\:{\rm d}s$$ is a martingale with respect to the filtration $$\mathcal F^{Y_n}$$ generated by $$Y_n$$. But why is $$\xi_n(t)-\int_0^t\varphi_n(s)\:{\rm d}s$$ still a $$\mathcal F^{Y_n}$$-martingale? Since they seem to conclude the proof with Theorem 8.2 (c), it seems like they are assuming this (since in Theorem 8.2, $$(\xi_n,\varphi_n)\in\hat{\mathcal A}_n$$)
• The printing of this book is often a bit hard to read, especially with $t$s in subscripts. Would question $1$ be resolved if you had $\phi_n(t) = g_n(Y_n(t)) \chi_{\tau_n \geq t}$? – Rhys Steele Apr 24 at 8:21
• I've looked in my print copy and that's definitely a $t$ and not a $1$. – Rhys Steele Apr 24 at 12:27
• @RhysSteele The funny thing is, I've ran text recognition to be sure and it said it's a $1$. – 0xbadf00d Apr 25 at 18:52

I think this all stems from you misreading the subscript in the definition of $$\varphi_n$$ (very understandable, it's almost impossible to discern subscripts in scanned copies of this book). The correct definition of $$\varphi_n(t)$$ is $$\varphi_n(t) = g_n(Y_n(t)) \chi_{\{\tau_n > t\}}.$$ The reason you want this is that it gives $$\xi_n(t) - \int_0^t \varphi_n(s) ds$$ the form of a stopped martingale. You've realised that $$M_n(t) := f_n(Y_n(t))-\int_0^tg_n(Y_n(s))\:{\rm d}s$$ is a martingale. Now notice that with correct definition of $$\varphi_n$$, we have $$\xi_n(t) - \int_0^t \varphi_n(s) ds = M_n(t)^{\tau_n} := M_n(t \wedge \tau_n)$$ so $$\xi_n(t) - \int_0^t \varphi_n(s) ds$$ is at least a local martingale for your filtration. Now you should be able to just use the usual tricks to see that this local martingale is a true martingale.
• Thank you for your answer. This makes more sense now. Just one subtlety: Shouldn't we define $$\tau_n:=\inf\left\{t>0:\left(\int_0^t\left|g_n(Y_n(s))\right|^2\:{\rm d}s\right)^{\frac12}>\sqrt t(\left\|g\right\|_\infty+1)\right\}$$ instead of the definition of $\tau_n$ given in the book? I guess this doesn't matter in the end, but $\tau_d$ should be defined with $$\sup_{y\in G_n}|g_n(y)|\le\sup_{y\in G_n}|g_n(y)-(\pi_ng)(y)|+\sup_{y\in G_n}|(\pi_ng)(y)|\le1+\left\|g\right\|_\infty$$ for $n$ large enough in mind. – 0xbadf00d Apr 25 at 19:00
• And regarding the true martingale property: Would your attempt be to show boundedness? We should have $$|M_n(\tau_n\wedge t)|\le\sup_{n\in\mathbb N}\left\|f_n\right\|_\infty+\sqrt t\left(\int_0^{\tau_n\:\wedge\:t}\left|g_n(Y_n(s))\right|^2\:{\rm d}s\right)^{\frac12}\le\underbrace{\sqrt t\sqrt{\tau_n\wedge t}}_{\le\:t}(1+\left\|g\right\|_\infty).$$ Is that sufficient? – 0xbadf00d Apr 25 at 19:22
• I don't have time to go through the proof to look for the answer to the first question but it really shouldn't matter. The two stopping times ought to have roughly the same behaviour and I would trust Ethier and Kurtz to have a good reason for their choice. For your second question, you've shown that on bounded time intervals the local martingale $M_n(t \wedge \tau_n)$ is uniformly bounded. This is a standard criterion for a local martingale to be a true martingale (remove the localising sequence appearing in the martingale property using the DCT). – Rhys Steele May 10 at 13:44