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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.?

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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.

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  • $\begingroup$ Are those g and h necessarily measurable though? The conditions on g and h mean they have to be measurable. $\endgroup$ – James Baxter Dec 25 '18 at 13:52
  • $\begingroup$ @JamesBaxter They are, since they agree with zero, respectively identity, function outside a set of measure zero. $\endgroup$ – Wojowu Dec 25 '18 at 13:53

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