I have two functions, and I want to combine them to define a certain function.


Suppose for every fixed $e$ in $(0, ∞)$, we have a function $g\_e (x): \mathbb{R} \to [0,\infty]$ that is well defined a.e. and right continuous, and for every fixed x in R, we have a function $h\_x (e): (0, ∞) \to [0, ∞]$ that is well defined a.e. and monotone increasing. Further, whenever $e$ and $x$ are such that they are both defined, $g\_e (x) = h\_x (e)$.


Can we canonically define some $f: \mathbb{R} \times (0, \infty)\to [0,\infty]$ such that it is well defined a.e. and $f(x, e) = g\_e (x) = h\_x (e)$ a.e.?



Given these conditions, can a function be defined that is well defined a.e.?
At least assuming the continuum hypothesis, the answer is *no*. Let $\prec$ be a well-order of $\mathbb R$ of order type $\omega\_1$. Define $g\_e(x)=0$ for $x\succ e$ (and $g\_e(x)$ not defined otherwise) and $h\_x(e)=e$ if $e\succ x$ (and $h\_x(e)$ not defined otherwise). Observe that both $g\_e,h\_x$ are defined everywhere except at a countable, hence null, set (this is why we need the well-order to have order type $\omega\_1$).


Then there is no pair $(x,e)$ for which both $g\_e(x),h\_x(e)$ are both defined, so the last condition is satisfied vacuously, but there is no function which is equal to both of those numbers simulataneously on any nonempty set.


