I am attempting to resolve a question I could not tackle in some project which I completed years ago in graduate school. In C.W. Gardners 'Handbook of Stochastic Methods, Third Ed.' it is claimed that the so-called continuity condition [for developing the so-called differential Chapman–Kolmogorov equation], \begin{equation}\tag{A.8} \begin{split} \lim_{\Delta t \to \infty} \frac{1}{\Delta t}\int_{|x-z|>\varepsilon} P(x,t+\Delta t|z,t ) dx = 0 \end{split} \end{equation}
is implied by the fulfillment of the Lindeberg condition. My best attempt to make the connection:
Given $|x-z|>\varepsilon$,
\begin{equation*} \begin{aligned}[b] \int_{|x-z|>\varepsilon} P(x,t+\Delta t|z,t ) d\textbf{x} &=\int_{|x-z|>\varepsilon} \frac{|\textbf{x}-\textbf{z}|^{2}}{|\textbf{x}-\textbf{z}|^{2}}P(\textbf{x},t+\Delta t|\textbf{z},t ) d\textbf{x} = 0 \\ &\leq \frac{1}{\varepsilon^{2}}\int_{|x-z|>\varepsilon}|\textbf{x}-\textbf{z}|^{2}P(\textbf{x},t+\Delta t|\textbf{z},t ) d\textbf{x} = 0 \\ & \qquad \qquad \qquad \qquad \vdots\\ &= \frac{1}{n \sigma^2}\sum_{i=1}^nE\left[|x_i-x_{i-1}|^{2}I_{|x_i-x_{i-1}|>\sqrt{n}\sigma \varepsilon}\right] \quad \rightarrow \quad 0 \end{aligned} \label{eq:1} \end{equation*}
Where the last line is the more common form of the Lindeberg condition. The vertical dots indicate my point of confusion.I am unsure how to go about relating the integral to a stochastic expectation value. My background is in physics not math, but I did take a stochastic course. Does this require something along the lines of a Martingale expectation?
*Update: This post is closely related, so much I am tempted to call this a duplicate. The received answer prove the continuity condition (there stated to be the equivalent of the Lindeberg) goes to zero via the Markov inequality. Yet relating this to the more usual form of Lindebergs condition or showing that the given proof implies a normal distribution was seemingly overlooked.
**Update 2: This paper and ch 3 of this book (preview of ch. 3 available on google books) shed some light on the situation. Its not trivial and I throw in the towel. But if anyone does think of an eloquent and short way to make the connection it is welcome.
