# Lipschitz continuity of an integral

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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The last displayed line should be $|P(x'',B)-\cdots|$? It's true if the measure of $B$ or $B^c$ is finite. – Davide Giraudo May 21 '12 at 15:38
@DavideGiraudo: thanks, that was a typo. Yeah, that's true if $B^c$ is of finite measure because $P(x,E) \equiv 1$. But I wonder if there is a counterexample for some $\mu(B) = \infty$ – Ilya May 21 '12 at 15:44
@Norbert I guess the OP meant the Lipschitz condition on compact subsets. – Davide Giraudo May 21 '12 at 18:14
@DavideGiraudo, if $\mu(B)=+\infty$, the last inequality is allways holds, isn't it? – Norbert May 21 '12 at 18:24
I agree, but the problem is to determine whether the map $x\mapsto P(x,B)$ is Lipschitz continuous over compact sets when $0<\mu(B)<\infty$. I agree that it's ambiguous. – Davide Giraudo May 21 '12 at 18:26

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 $$P(x,B) = \int_1^{1/\sqrt{|x|}}\frac1{y^2}-|x|\,dy = 1-2\sqrt{|x|}+|x|.$$