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What are the conditions in the definition of Caratheodory functions?

Let $f\colon T\times X\to \mathbb{R}$, where $T\subset \mathbb{R}$ and $X$ is real Banach space.

1) $f(\cdot,x)$ is measurable for every $x\in X$,

2) $f(t,\cdot)$ is continuous for almost every $t\in T$,

3) $f(\cdot,x)\in L(T)$.

Are there any more?

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  • $\begingroup$ Have you found the answer? I also have the same question :) $\endgroup$
    – ABIM
    Commented Jun 20, 2018 at 12:54

3 Answers 3

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The answer, as formulated for general topological spaces, is at the beginning of this paper :)

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On page 3 of this paper by Longo, Novo and Obaya, you can find the definition of a strong Carathéodory function.

You can compare that with the conditions given on this section of the Wikipedia article for the Carathéodory existence theorem.

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Although this is an old question, I think the following is still relevant. From the posting of the OP it seems that at the time of the posting the author might have been interested in applications to ODE's where integrability (condition 3) is used to prove existence of absolutely continuous solution to differential equations of the form $$\dot{x}(t)=f(t,x(t))$$ where $f$ is defined on a domain $D\subset\mathbb{R}\times\mathbb{R}^n$, where for every bounded closed interval $I\subset \mathbb{R}$ and compact set $U\subset \mathbb{R}^d$ such that $I\times U\subset D$; $f:I\times U\rightarrow\mathbb{R}^d$ is continuous on $U$ for each $t\in I$ fixed, and measurable on $I$ for each $x\in U$.

There is however other applications of Caratheodory functions in economics and statatistics for which condition (3) is not even defined. This is from Aliprantis, C. D, and Border, K. C., Infinite Dimensional Analysis, Third edition, Springer-Verlag, Berlin Heidelberg, 2006.

Definition Let $(S , \Sigma)$ be a measurable space, and let $X$ and $Y$ be topological spaces. A function $f : S \times X \rightarrow Y$ is a Carathéodory function if:

  1. for each $x\in X$, the function $f^x = f(·,x): S \rightarrow Y$ is $(\Sigma,\mathscr{B}_Y)$-measurable; and
  2. for each $s\in S$, the function $f_s = f(s,·): X \rightarrow Y$ is continuous.

Carathéodory functions have the virtue of being jointly measurable in many important cases.

Lemma (Carathéodory functions are jointly measurable) Let $(S, \Sigma)$ be a measurable space, $X$ a separable metrizable space, and $Y$ a metrizable space. Then every Carathéodory function $f: S \times X \rightarrow Y$ is jointly measurable.

Proof: Let $d$ and $\rho$ be metrics on $X$ and $Y$ respectively compatible with their topologies. Let $(x_n:n\in\mathbb{N})\subset X$ be dense on $X$. For any $s\in S$, the continuity of $f(s,\cdot)$ implies that for any closet set $F\subset Y$ ans $x\in X$, $f(s,x)\in F$ iff for each $n\in\mathbb{N}$ there is some $x_m$ with $d(x,x_m)< 1/n$ and $\rho(f(s,x_m),F)<1/n)$. Hence \begin{align} f^{-1}(F)=\bigcap^\infty_{n=1}\bigcup^\infty_{m=1}\Big\{s\in S: f(s:x_m)\in F^{1/n}\}\times B(x_n;1/n) \end{align} where $F^{1/n}:=\{y\in Y: \rho(y,F) <1/n\}$. As $f$ is measurable in $s$ and $F^{1/n}$ is open (and hence Borel), each set $\{s\in S: f(s:x_m)\in F^{1/n}\}$ is measurable. Consequently $f^{-1}(F)$ is measurable. $\blacksquare$

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