Necessary topological conditions for all Caratheodory functions to be measurable If $(S,\Sigma)$ is a measurable space, and $X$ and $Y$ are topological spaces, then a function $f:S \times X \rightarrow Y$ is called a "Caratheodory function" if $s \mapsto f(s,x)$ is measurable (with respect to the Borel $\sigma$-algebra on $Y$) for each $x$ in $X$ and $x \mapsto f(s,x)$ is continuous for each $s$ in $S$. It is well-known that every Caratheodory function is measurable with respect to the product $\sigma$-algebra $\Sigma \otimes \mathcal{B}_X$ on $S \times X$ and the Borel $\sigma$-algebra on $Y$ when $X$ and $Y$ are metrizable spaces and $X$ is separable.
I would like to examine what happens when $X$ and $Y$ are not necessarily metrizable. I have in mind the particular example that every Caratheodory function is measurable if $Y$ is $\mathbb{R}^n$ and $X$ is $\mathbb{R}^+_\ell$ or $\mathbb{R}^+_u$ (i.e. $[0,\infty)$ equipped with the lower or upper limit topologies), for example by the construction used in this question.
I can prove every Caratheodory function is measurable under the assumptions (strictly weaker than $X$ is separable metrizable and $Y$ is metrizable) that $X$ is second countable and $Y$ is perfectly normal. To see this, suppose $f$ is a Caratheodory function; it suffices to show $f^{-1}(F)$ is measurable for any closed set $F \subset Y$. Since $Y$ is perfectly normal, every closed subset is $G_\delta$, so there is a countable sequence of open sets $V_1,V_2,\dots$ whose intersection is $F$; then using normality we can construct open sets $W_1,W_2,\dots$ such that $F \subset W_i \subset \overline{W_i} \subset V_i$ for each $i$ (incidentally, the existence for every closed set $F$ a countable collection of open sets, each containing $F$ and for which the intersection of their closures is $F$ implies the space is perfectly normal, and therefore is an equivalent characterization of perfect normality; see this question). Then if $\{U_1,U_2,\dots\}$ is a countable base for $X$ and $\mathcal{X}$ is a countable dense subset of $X$ (which can be obtained, for example, by picking an element from each $U_i$), we have
$$f^{-1}(F) = \bigcap_{i\in \mathbb{N}} \,\bigcup_{j \in \mathbb{N}} \, \bigcap_{x \in \mathcal{X} \cap U_j}\{s \in S\,|\, f(s,x) \in \overline{W_i}\} \times U_j,$$
which shows $f^{-1}(F)$ is in $\Sigma \otimes \mathcal{B}_X$.
Frustratingly, this does not cover the case I mentioned earlier, as the lower/upper limit topologies on $\mathbb{R}$ are not second countable (they are separable and first countable, but not second countable). Are there ways to relax the conditions on the topologies of $X$ and $Y$ to account for this case? Is it possible to give nice necessary conditions on the topologies of $X$ and $Y$ for all Caratheodory functions to be measurable?
 A: Well, after thinking some more, I've found weaker topological conditions on $X$ that cover these other cases, so I'm answering my own question if for no other reason than to put down my thoughts. I'm not totally satisfied with this answer, but it at least works.
If $X$ is a topological space, let $\mathcal{M}(X,X)$ denote the set of all (Borel) measurable functions from $X$ to $X$, which we equip with the topology of pointwise convergence, and let $B_f(X)$ denote the subset of measurable functions $h$ whose images are finite sets. I'm going to say a topological space $X$ "has sequentially-approximable self-maps" if the smallest sequentially closed set in $\mathcal{M}(X,X)$ containing $B_f(X)$ is, in fact, all of $\mathcal{M}(X,X)$ (i.e. the transfinite sequential closure of $B_f(X)$ is $\mathcal{M}(X,X)$).
Note that this is equivalent to saying that just the identity map is in the transfinite sequential closure, as if $f:X \rightarrow X$ is any other measurable function, then $h \circ f$ is in $B_f(X)$ if $h$ is, and if we let $\{B_f^\alpha(X)\}_{\alpha \in \text{Ord}}$ denote the transfinite sequence of sequential closures of $B_f(X)$, then transfinite induction shows that $h \circ f$ is in $B_f^\alpha(X)$ if $h$ is (as if $h_n \rightarrow h$ pointwise, then $h_n \circ f \rightarrow h \circ f$ pointwise), and so if the identity map is in the transfinite sequential closure of $B_f(X)$ (i.e. $B_f^{\omega_1}(X)$, where $\omega_1$ is the first uncountable cardinal), then so is $\text{id} \circ f = f$.
Now, I claim that if $(S, \Sigma)$ is a measurable space, $X$ is a topological space which has sequentially-approximable self-maps, and $Y$ is a perfectly normal topological space, then every Caratheodory function $f: S \times X \rightarrow Y$ is measurable. To see this, for any $h:X \rightarrow X$, let $f_h(s,x) = f(s,h(x))$. We will show that $f_h$ is measurable for all $h$ in $B_f^\alpha(X)$ by transfinite induction. For the base case, we note that if $F$ is any Borel subset of $Y$ and $h$ is any function in $B_f(X)$, then
$$f_h^{-1}(F) = \bigcup_{x \in \text{im } h} \{s\in S\,|\, f(s,x) \in F\} \times h^{-1}(B_x),$$
where $B_x$ is any Borel subset of $X$ whose intersection with $\text{im }h$ is precisely the points in $\text{im }h$ which are topologically indistinguishable from $x$ (i.e. we can take $B_x = \{x\}$ if singletons are Borel in $X$; otherwise for each $y$ in $\text{im }h$ which is distinguishable from $x$, there is either an open or closed set containing $x$ and not $y$, so the intersection of these sets for every $y \in \text{im }h$ distinguishable from $x$ gives a set satsifying the requirements of $B_x$). This expresses $f_h^{-1}(F)$ as a finite union of measurable sets, since $s \mapsto f(s,x)$ is measurable for each $x$, $\text{im }h$ is finite, and $f(s,x) \in F$ if and only if $f(s,y) \in F$ when $x$ and $y$ are topologically indistinguishable, as $x \mapsto f(s,x)$ is continuous for each $s$, and therefore must send indistinguishable points to indistinguishable points, and the set $F$ is Borel in $Y$, and therefore contains or is disjoint from each indistinguishability class. Thus, $f_h$ is measurable, and since $h$ was arbitrary, we, in fact, have that $f_h$ is measurable for all $h$ in $B_f^0(X)$. Now, suppose $\alpha$ is a successor ordinal, $f_h$ is measurable for all $h$ in $B_f^{\alpha-1}(X)$, and $h$ is in $B_f^\alpha(X)$. Then there is a sequence $\{h_n\}_{n \in \mathbb{N}}$ in $B_f^{\alpha-1}(X)$ such that $h_n \rightarrow h$ pointwise. Then, since $x \mapsto f(s,x)$ is continuous for each $s$ in $S$, we have that $f(s,h_n(x)) \rightarrow f(s,h(x))$ for each $s$ in $S$ and $x$ in $X$. Thus, $f_{h_n} \rightarrow f_h$ pointwise, and since $Y$ is perfectly normal, this means $f_h$ is measurable (see Cristian Paris's answer and my comment here). On the other hand, if $\alpha$ is a limit ordinal, and we assume that $f_h$ is measurable if $h$ is in $B_f^{\beta}(X)$ for any $\beta < \alpha$, if $h$ is in $B_f^{\alpha}(X)$, then $h$ is in $B_f^{\beta}(X)$ for some $\beta < \alpha$, and so $f_h$ is measurable. Thus, the induction is complete, and so since the identity map is in $B^{\omega_1}_f(X)$, we have $f_{\text{id}} = f$ is measurable, which is what we wanted.
Now I want to note that this encompasses all of the situations I discussed in my question, i.e. that second countable spaces necessarily have sequentially-approximable self-maps, but that so do $\mathbb{R}_\ell^+$ and $\mathbb{R}_u^+$. So first we show that second countable spaces necessarily have sequentially-approximable self-maps. Suppose $\{U_1,U_2,\dots\}$ is a countable base for a topological space $X$. Then for every subset $S \subset \{1,\dots,n\}$, let
$$U^n_S = \left(\bigcap_{i \in S} U_i \right) \cap \left(\bigcap_{i \in S^c} U_i^c \right),$$
where complementation is done in $\{1,\dots,n\}$ or $X$ accordingly. Then, for each $n$, $\{U^n_S\}_{S \subset \{1,\dots,n\}}$ is a (finite) Borel partition of $X$; we will define functions which are constant on these sets which converge to the identity map. For each natural $n$ and set $S \subset \{1,\dots,n\}$ for which $U^n_S$ is nonempty, let $x^n_S$ be a point in $U^n_S$ (any point suffices). Then let $f_n:X \rightarrow X$ be the function which is constant and equal to $x^n_S$ on $U^n_S$ for each $S \subset \{1,\dots,n\}$ (which is valid since $\{U^n_S\}_{S \subset \{1,\dots,n\}}$ is a partition of $X$). Note that $f_n$ has finite image (its image is $\{x^n_S\}_{S \subset \{1,\dots,n\}}$), and that, for any $F$, $f_n^{-1}(F)$ is a union of some subcollection of $\{U^n_S\}_{S \subset \{1,\dots,n\}}$, which means $f_n$ is measurable. Thus, we will have shown $X$ has sequentially-approximable self-maps if $f_n \rightarrow \text{id}$ pointwise. To see this, fix any point $x$ in $X$ and any open set $V$ containing it; we will show $f_n(x)$ is in $V$ for sufficiently large $n$, which shows that $f_n(x) \rightarrow x$, which is what we want. There's an $i$ such that $x \in U_i \subset V$ since $\{U_1,U_2,\dots\}$ is a base for the topology of $X$. Thus, for $n \geq i$, we have that $f_n(x) = x^n_S \in U^n_S$ where $S$ is the subset of $\{1,\dots,n\}$ such that $x$ is in $U^n_S$. Since $U^n_S$ must be a subset of $U_i$ (since $n \geq i$, it must be a subset of $U_i$ or of $U_i^c$, and since it contains $x$, an element of $U_i$, it must be a subset of $U_i$), we have that $f_n(x)$ is in $U_i$ for all $n \geq i$, so $f_n(x)$ is in $V$ for all $n \geq i$. Thus, $f_n(x) \rightarrow x$ for all $x$ in $X$, and so we are done.
Finally, we show that $\mathbb{R}_\ell^+$ and $\mathbb{R}_u^+$ have sequentially-approximable self-maps as well. For each natural $n$, let $f_n^\ell$ be the function which is equal to $k/n$ on the interval $[(k-1)/n,k/n)$ for $k$ in $\{1,\dots,n^2\}$ and is zero on $[n,\infty)$, and similarly let $f_n^u$ be the function which is equal to $(k-1)/n$ on the interval $((k-1)/n,k/n]$ for $k$ in $\{1,\dots,n^2\}$ and is zero on $\{0\} \cup (n,\infty)$. Then $f_n^\ell$ is in $B_f(\mathbb{R}_{\ell}^+)$ and $f_n^u$ is in $B_f(\mathbb{R}_u^+)$ for every $n$, and it is easy to see that $f_n^\ell \rightarrow \text{id}$ pointwise in $\mathbb{R}_{\ell}^+$ and $f_n^u \rightarrow \text{id}$ pointwise in $\mathbb{R}_u^+$.
I'm still not completely satisfied with this answer, as it somehow feels like it was reverse engineered to account for the $\mathbb{R}_\ell^+$ and $\mathbb{R}_u^+$ cases. Nonetheless, it does give sufficiently weak topological conditions to cover the cases I was concerned about.
