Based on here: Is the refutation of the following claim correct? I'm trying to see if I understand antkam correctly and if antkam is right.
Claim:
It is possible to have $F_{i}(t)$s be unknown CDFs and $G_{i}(t)$s be known pdfs s.t.
$$F_{1}'(t)=\dfrac{G_{1}(t)}{(1-F_{2})F_{3}}$$ $$F_{2}'(t)=\dfrac{G_{2}(t)}{(1-F_{3})F_{1}}$$ $$F_{3}'(t)=\dfrac{G_{3}(t)}{(1-F_{1})F_{2}}$$
Refutation:
We must have all denominators less than 1 or all denominators equal to 1.
We rule out the latter because this contradicts the non-decreasing property of CDFs. It remains to rule out the former.
$$\frac{G_{1}(t)}{(1-F_{2})F_{3}} > G_{1}(t) \iff G_1(t) = 0 \iff 1 = (1-F_{2})F_{3}$$
We cannot have that $$\int_{\mathbb R} \dfrac{G_{1}(t)}{(1-F_{2})F_{3}} dt > \int_{\mathbb R} G_{1}(t) dt$$ because $$1 = \int_{\mathbb R} \dfrac{G_{1}(t)}{(1-F_{2})F_{3}} dt = \int_{\mathbb R} G_{1}(t) dt $$, but we do have that from 3 and because here.
For #1, must we? If $F_3(t_a) = 1$, then I was thinking $G_2(t_a) = 0$?
For #2, what if the union of such t's has measure zero? Do we need to use Lebesgue measure?
Yes. Suppose that $F_3(a) = 1$; then (as you noted in the comments here) $G_2(a) = 0$ in order to have any hope of these equations making sense. In fact, this implies that $F_3(t) = 1$ and $G_2(t) = 0$ for all $t \geq a$. Having $G_2(t) = 0$ on $[a, \infty)$ implies that $F_2(t) = 1$ on $[a, \infty)$ since we must have $\lim_{t \to \infty} F_2(t) = 1$ and $F_2(t) - F_2(a) = \int_a^t G_2(s) \, \textrm d s$. Then we repeat the above argument to show that $F_2(t) = 1$ on $[a, \infty)$ implies that $F_1(t) = 1$ on $[a, \infty)$.
If by "such $t$'s" you mean $\{t :F_i(t) = 1\}$, then this set is either empty or of infinite measure. Since $F_i(t)$ is a nondecreasing function bounded above by $1$, then $F_i(a) = 1$ implies that $F_i(t) = 1$ for all $t \geq a$.