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For a measurable space $(E,\mathcal E)$ and a Markov kernel $P:E\times \mathcal E\to[0,1]$ there is a unique homogeneous Markov chain $X$. The first return time is defined as $$ \tau_A = \inf\{k\geq 1:X_k\in A\}. $$ Denote $L(x,A) = \mathbb P_x[\tau_A<\infty]$ for all $x\in E$ and $A\in \mathcal E$.

I have the following question: is there $\delta(P)>0$ such that for all Markov kernels $Q$ satisfying $$ \|P-Q\| = \sup\limits_{\begin{align}x&\in E\\B&\in \mathcal E\end{align}}|P(x,B)-Q(x,B)|<\delta $$ the property $L_P(x,A) =1$ for all $x\in E$ and some $A$ implies $L_Q(x,A) = 1$ for all $x\in E$?

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That's not the correct definition of recurrence. You don't want the expected return time to be $1$, you want the return time to be finite with probability $1$. Also you'll want to assume $x \in A$, not just $x \in E$. – Robert Israel Feb 17 '12 at 20:21
@RobertIsrael: thanks for fixing the typo - but in fact I meant that $L(x,A)$ has to be one for all $x\in E$. If this is not the definition of recurrence, I am sorry - but I meant that when asked the question. I edited the question to make it clear. – S.D. Feb 17 '12 at 22:30

Consider a two-state Markov chain. For $P$ the transition matrix is $\pmatrix{1 & 0\cr 0 & 1\cr}$, while for $Q$ it is $\pmatrix{ 1-\delta & \delta\cr 0 & 1\cr}$. Then state $1$ is recurrent for $P$ but transient for $Q$.

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