- x| > K} \frac{p(h, x, x')}{h}.\end{gathered}$$

Rewrite now the last summand on the right side of : $$\begin{gathered}
\label{AE1}
\sum_{| x' - x| > K} \frac{p(h, x, x')}{h} = \frac{1 - p(h, x, x)}{h} - \sum_{0 < | x' - x| \leqslant K}  \frac{p(h, x, x')}{h} \\
= \left( \frac{1 - p(h, x, x)}{h} + a(x, x) \right) - \sum_{0 < | x' - x| \leqslant K} \left(  \frac{p(h, x, x')}{h} - a(x, x') \right)\\
-  a(x,x) -  \sum_{0 < | y - x| \leqslant K} a(x, y)\end{gathered}$$

Substituting $a(x,x)=-\sum_{y\neq x}a(x,y)$ in , we finally estimate :

$$\begin{gathered}
\sum_{x'\neq x} \left|  \frac{p(h, x, x')}{h} - a(x, x') \right| \leqslant 2 \sum_{0 < | x' - x| \leqslant K} \left|  \frac{p(h, x, x')}{h} - a(x, x') \right|  + 2 \sum_{| x' - x| > K} a(x, x')\\
+  \left|  \frac{1-p(h, x, x)}{h} + a(x, x) \right|.\end{gathered}$$

Here the last summand tends to zero when $ h \to 0 $ by virtue of the relation ; the second sum could be made arbitrary small by choosing sufficiently large $K$ (since the series $\sum_{x'}a(x, x')$ converges absolutely); with the fixed $K$ the first sum tends to zero when $ h \to 0 