[proofplan] We combine measurable simple approximation with the continuity part of the Caratheodory hypothesis. First we regard $(u,v)$ as one measurable map into the finite-dimensional state space $\mathbb R^{1+n}$ and approximate it pointwise by countably valued measurable maps. For each approximation, the composed function is measurable because it is assembled from measurable fixed-state sections of $f$. Continuity in the state variables gives pointwise convergence outside the exceptional null set, and the value on that null set does not affect Lebesgue measurability. [/proofplan]

[step:Package the state variables into one measurable map] Let $Z:=\mathbb R\times\mathbb R^n$, equipped with its Euclidean Borel $\sigma$-algebra. Define the map \begin{align*} w:U\to Z,\qquad w(x)=(u(x),v(x)). \end{align*} Since $u:U\to\mathbb R$ and $v:U\to\mathbb R^n$ are Lebesgue measurable, each coordinate function of $w$ is Lebesgue measurable. Because the Borel $\sigma$-algebra of $Z=\mathbb R^{1+n}$ is generated by coordinate half-spaces, $w$ is Lebesgue measurable as a map from $U$ into $Z$. Let $N\subset U$ be the exceptional set from the Caratheodory hypothesis, so that $\mathcal L^n(N)=0$ and, for every $x\in U\setminus N$, the map \begin{align*} F_x:Z\to(-\infty,\infty],\qquad F_x(z)=f(x,z) \end{align*} is continuous. Here $f(x,z)$ denotes $f(x,s,\xi)$ when $z=(s,\xi)\in \mathbb R\times\mathbb R^n$. [/step]

[step:Choose countably valued measurable approximations to the state map]For each $k\in\mathbb N$, partition $Z=\mathbb R^{1+n}$ into countably many Borel sets $(Q_{k,j})_{j=1}^{\infty}$ of diameter at most $2^{-k}$, with each nonempty $Q_{k,j}$ assigned a point $z_{k,j}\in Q_{k,j}$. Define \begin{align*} w_k:U\to Z,\qquad w_k(x)=z_{k,j}\text{ whenever }w(x)\in Q_{k,j}. \end{align*} Each set $w^{-1}(Q_{k,j})$ is Lebesgue measurable because $w$ is measurable and $Q_{k,j}$ is Borel. Hence each $w_k$ is a countably valued Lebesgue measurable map. Moreover, for every $x\in U$ and every $k\in\mathbb N$, the points $w_k(x)$ and $w(x)$ lie in the same set $Q_{k,j}$, so \begin{align*} |w_k(x)-w(x)|\le 2^{-k}. \end{align*} Therefore $w_k(x)\to w(x)$ in $Z$ for every $x\in U$.[/step]

[guided]The reason for introducing the maps $w_k$ is that the function $x\mapsto f(x,w(x))$ is difficult to test directly: both the $x$ variable and the state variable $w(x)$ move at the same time. We freeze the state variable on measurable pieces. For each $k\in\mathbb N$, choose a countable Borel partition $(Q_{k,j})_{j=1}^{\infty}$ of $Z=\mathbb R^{1+n}$ such that every set in the partition has Euclidean diameter at most $2^{-k}$. This can be obtained, for instance, by intersecting the dyadic cubes of side length small enough with annuli and then enumerating the resulting countable family. For each nonempty cell $Q_{k,j}$, choose one representative point $z_{k,j}\in Q_{k,j}$. Define \begin{align*} w_k:U\to Z,\qquad w_k(x)=z_{k,j}\text{ if }w(x)\in Q_{k,j}. \end{align*} This definition is unambiguous because the sets $Q_{k,j}$ form a partition. Since $w:U\to Z$ is measurable and each $Q_{k,j}$ is Borel, the set $w^{-1}(Q_{k,j})$ is Lebesgue measurable in $U$. Thus $w_k$ is measurable: the preimage of any subset of its countable range is a countable union of sets of the form $w^{-1}(Q_{k,j})$. Finally, if $w(x)\in Q_{k,j}$, then $w_k(x)=z_{k,j}\in Q_{k,j}$ as well. Since $Q_{k,j}$ has diameter at most $2^{-k}$, we get \begin{align*} |w_k(x)-w(x)|\le 2^{-k}. \end{align*} Letting $k\to\infty$ gives $w_k(x)\to w(x)$ for every $x\in U$. This is exactly the approximation needed to use continuity of the state map $z\mapsto f(x,z)$.[/guided]

[step:Show each approximating composition is measurable] For each $k\in\mathbb N$, define \begin{align*} g_k:U\to(-\infty,\infty],\qquad g_k(x)=f(x,w_k(x)). \end{align*} For fixed $k$, the map $w_k$ has countable range contained in $\{z_{k,j}:j\in\mathbb N\}$. For every $a\in\mathbb R$, the superlevel set of $g_k$ is \begin{align*} \{x\in U:g_k(x)>a\}=\bigcup_{j=1}^{\infty}\bigl(w^{-1}(Q_{k,j})\cap\{x\in U:f(x,z_{k,j})>a\}\bigr). \end{align*} The set $w^{-1}(Q_{k,j})$ is Lebesgue measurable, and the set $\{x\in U:f(x,z_{k,j})>a\}$ is Lebesgue measurable by the fixed-state measurability part of the Caratheodory hypothesis. Therefore $\{x\in U:g_k(x)>a\}$ is Lebesgue measurable for every $a\in\mathbb R$, so $g_k$ is Lebesgue measurable as an extended-real-valued function. [/step]

[step:Pass to the pointwise limit off the exceptional set] Define \begin{align*} g:U\to(-\infty,\infty],\qquad g(x)=f(x,w(x)). \end{align*} If $x\in U\setminus N$, then $F_x:Z\to(-\infty,\infty]$ is continuous and $w_k(x)\to w(x)$ in $Z$. Hence \begin{align*} g_k(x)=F_x(w_k(x))\to F_x(w(x))=g(x) \end{align*} in $(-\infty,\infty]$ for every $x\in U\setminus N$. Define the auxiliary function \begin{align*} \tilde g:U\to(-\infty,\infty],\qquad \tilde g(x)=\limsup_{k\to\infty} g_k(x). \end{align*} Since each $g_k$ is measurable, $\tilde g$ is measurable: for every $a\in\mathbb R$, \begin{align*} \{x\in U:\tilde g(x)>a\}=\bigcap_{m=1}^{\infty}\bigcup_{k\ge m}\{x\in U:g_k(x)>a\}, \end{align*} which is Lebesgue measurable. The preceding convergence shows that $\tilde g(x)=g(x)$ for every $x\in U\setminus N$. [/step]

[step:Use completeness of Lebesgue measure to restore the null set] It remains to pass from $\tilde g$ to $g$ on the null set $N$. For every $a\in\mathbb R$, \begin{align*} \{x\in U:g(x)>a\}=\bigl(\{x\in U\setminus N:\tilde g(x)>a\}\bigr)\cup\bigl(\{x\in N:g(x)>a\}\bigr). \end{align*} The first set is Lebesgue measurable because $\tilde g$ is Lebesgue measurable and $U\setminus N$ is Lebesgue measurable. The second set is a subset of $N$, hence Lebesgue measurable because [Lebesgue measure is complete](/theorems/4909) and $\mathcal L^n(N)=0$. Therefore $\{x\in U:g(x)>a\}$ is Lebesgue measurable for every $a\in\mathbb R$. Thus \begin{align*} x\mapsto g(x)=f(x,u(x),v(x)) \end{align*} is Lebesgue measurable on $U$. This is the desired conclusion. [/step]