# Interchanging supremum and integral (another question)

Suppose that $\varphi$ is a smooth $\mathcal{K}_{\infty}$-function and $\bar{\textbf{B}}$ is the unit ball set in $\mathbb{R}^{n}$. Let $x : \mathbb{R}_{\geq 0} \times \mathbb{R}^{n} \to \mathbb{R}^{n}$ be such that $x(.,\xi)$ is differentiable for each $\xi$ and $x(t,.)$ is uniformly locally Lipschitz for each $t$. Also let $\omega : \mathbb{R}^{n} \to \mathbb{R}_{\geq 0}$ be a continuous function on $\mathbb{R}^{n}$ with $\omega(0) = 0$ and $\omega(x) \to \infty$ as $x \to \infty$. I'd like to know that whether the following holds $$\sup_{\xi \in \bar{\textbf{B}}} \int_{s=0}^{s=t}{\varphi(\omega(x(s,\xi)))ds} = \int_{s=0}^{s=t}{\sup_{\xi \in \bar{\textbf{B}}} \varphi(\omega(x(s,\xi)))ds}.$$ In other words, are the supremum and integral signs interchangable?

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What does "(another question)" in the title mean? – joriki May 4 '13 at 20:19

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter $t$; let $\mathcal{B}_X$ be a topological base $X$ and $\mathcal{B}_{T}$ a topological base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality