# A different Uniform Bound on a Sequence of Uniform Bounded Integrals

Let $m$ be a probability measure on $W = \mathbb{R}^m$, so that $m(W)=1$.

Consider a sequence $\{X_k\}_{k=1}^{\infty}$ of compact sets $X_k \subset X = \mathbb{R}^n$ such that $X_k \rightarrow X$.

Consider a locally bounded function $f: X \times W \rightarrow \mathbb{R}_{\geq 0}$.

We say that a function $\varphi: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ is of class $\mathcal{C}$ if it is continuous, strictly-increasing, zero-at-zero and unbounded.

Assume that there exists a class-$\mathcal{C}$ function $F: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ and a (uniform) $M \in \mathbb{R}_{>0}$ such that for any $k \in \mathbb{Z}_{\geq 1}$ we have

$$\sup_{x \in X_k} \int_W F(f(x,w)) m(dw) < M$$

Prove that there exists a concave class-$\mathcal{C}$ function $G$ and a (uniform) $M_G \in \mathbb{R}_{>0}$ such that for any $k \in \mathbb{Z}_{\geq 1}$ we have

$$\sup_{x \in X_k} \int_W G(f(x,w)) m(dw) < M_G$$

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Just curious, how do you define $X_k \to X$? –  Mercy Jul 4 '12 at 21:20
$\lim_{k\rightarrow \infty} X_k = X$. For instance: $X_k = k \mathbb{B}$ (closed ball of radius $k$). –  Adam Jul 4 '12 at 22:44
What is not clear with the limit $\lim_{k \rightarrow \infty} X_k = X$? If you are not familiar with that, you can try to prove the claim with $X_k := k \mathbb{B}$. –  Adam Jul 5 '12 at 0:03