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This question gives interesting insights on whenever uniform boundedness can be "iterated".

Let $\mu(\cdot)$ be a probability measure on the closed set $Z \subseteq \mathbb{R}^p$, so that $\int_Z \mu(dz) = 1$.

Consider a locally bounded function $\phi: X \times Y \times Z \rightarrow X$, with $X \subseteq \mathbb{R}^n$, $Y \subset \mathbb{R}^m$ compact, such that

$\forall z \in Z$ the map $(x,y) \mapsto \phi(x,y,z)$ is continuous;

$\forall (x,y) \in X \times Y$ the map $\phi(x,y,z)$ is measurable.

Let $\psi: X \times Y \rightarrow \mathbb{R}_{\geq 0}$ be a continuous function.

For any $x \in X$, define the function $\Psi_{x,\delta}: Z \rightarrow \mathbb{R}_{\geq 0}$ as follows. $$ \Psi_{x,\delta}(z) := \sup\{ \psi( \phi(\xi,y_1,z),y_2) \mid |x-\xi|\leq\delta, (y_1,y_2)\in Y^2 \} $$

Assume uniform boundedness: there exists $\delta>0$ such that $$ \int_Z \Psi_{x,\delta}(z) \mu(dz) < \infty $$ Induction: for any $x \in X$, define the function $\bar \Psi_{x,\delta}: Z^2 \rightarrow \mathbb{R}_{\geq 0}$ as follows.

$$ \bar \Psi_{x,\delta}(z_1,z_2) := \sup\{ \psi( \phi(\phi(\xi,y_1,z_1),y_2,z_2),y_3) \mid |x-\xi|\leq\delta, (y_1,y_2,y_3)\in Y^3 \} $$

Is it true that

$$ \int_{Z^2} \bar \Psi_{x,\delta}(z_1,z_2) \mu(dz_1) \mu(dz_2) < \infty $$

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Please don't shout. – Arturo Magidin May 14 '12 at 2:01

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