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Let $(E,d)$ be a metric space, $\mathscr E$ be its Borel $\sigma$-algebra and $\mu$ be a $\sigma$-finite measure on $(E,\mathscr E)$. Let the function $p:E\times E\to\mathbb R_+$ be non-negative and jointly measurabe: $p\in\mathscr E\otimes \mathscr E$. Let's assume that for any compact set $A\subset E$ there is a constant $\lambda_A$ such that $ |p(x'',y) - p(x',y)|\leq \lambda_A\cdot d(x',x'')\text{ for all }x',x''\in A,y\in E \tag{1} $ and let us assume that $ \int\limits_E p(x,y)\mu(\mathrm dy) = 1 \tag{2}\text{ for all }x\in E. $

Clearly, if $A$ is compact, then $P(x,B):=\int\limits_B p(x,y)\mu(\mathrm dy)$ is Lipschitz on $A$ whenever $\mu(B)<\infty$: $ |P(x'',B) - P(x',B)|\leq \lambda_A\cdot\mu(B)d(x'',x'). $ Does the Lipschitz continuity of $P(x,B)$ also hold on $A$ if $\mu(B)=\infty$?

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    @Davide isn't any function locally Lipschitz in discrete topology, since the only compact sets are finite sets?2012-05-22

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The Lipschitz continuity may fail if $\mu(B) = \infty$. Take for instance $E = \mathbb{R}$, $\mu$ to be the Lebesgue measure and define $p(x,y) = \max\{0,y^{-2}-|x|\}$ for $y >1$ and suitably for $y\le 1$ to have (1) and (2) satisfied. Let $B = [1,\infty)$. Then $P(\cdot,B)$ is not Lipschitz at $0$: $P(0,B) = 1$ and $P(x,B) = \int_1^{1/\sqrt{|x|}}\frac1{y^2}-|x|\,dy = 1-2\sqrt{|x|}+|x|.$