$$\int_{\mathbb R} f dt = 1 = \int_{\mathbb R} \frac f h dt \to h \equiv 1?$$
The context is probability, and I want to know if this can be generalised hoping to use as few properties of pdfs or cdfs as possible and to use as little measure theory as possible.
We have for all $t \in \mathbb R$,
- $f \in [0,1]$
- $h: = (1-F_2)F_3$
- $F_2'(t)$ and $F_3'(t)$ exist and
- $F_2(t), F_3(t), F_2'(t), F_3'(t) \in [0,1]$
- hence $h:= (1-F_2)F_3 \in [0,1]$
Case 0: (the original)
$$F_{1}'(t)=\dfrac{G_{1}(t)}{(1-F_{2})F_{3}}$$ $$F_{2}'(t)=\dfrac{G_{2}(t)}{(1-F_{3})F_{1}}$$ $$F_{3}'(t)=\dfrac{G_{3}(t)}{(1-F_{1})F_{2}}$$
where $F_{i}(t)$s are unknown CDFs, and $G_{i}(t)$s are known pdfs
Then $h \equiv 1 \equiv (1-F_{2})F_{3}=(1-F_{3})F_{1}=(1-F_{1})F_{2}$
Case 1: $h > 0$ and $h$ is integrable
Then $\frac 1 h \ge 1$.
I think it can be shown that if $\frac 1 h = c_1 \times 1_{A_1} + c_2 \times 1_{A_1}$ for $c_1, c_2 \ge 1$ and $A_1$ and $A_2$ is a partition of $\mathbb R$, then we must have $c_1 = c_2 = 1$.
Then we can extend to simple functions, nonnegative functions and integrable functions. Um, do we need $A_1, A_2$ to be $\mu$-measurable under some measure space $(\mathbb R, \Sigma, \mu)$ ? I was thinking Lebesgue outer measure but since that's not countably additive, perhaps we don't have extension from simple to nonnegative?
Case 2: $h > 0$ and $h$ is not integrable, i.e. $\int_{\mathbb R} h^{+} = \infty$ xor $\int_{\mathbb R} h^{-} = -\infty$
But $h \in [0,1]$ is bounded and hence integrable.
Case 3: $h$ can be $0$ Then choose $\frac 1 h = 5 \times 1_{A}$ where $\int_A f dt = \frac15$
but I'm guessing that the choice somehow defies cdf definitions? Or even if it doesn't, I think there's going to be a problem with dividing by indicator functions. I'm thinking something like $\int_{A^C} \frac{f}{1_A} dt$ with $\int_{A^C} f dt = 0$. Somewhere I came across something like $\int_A f dt = \int_A \frac f h dt = 1$ where $A \subseteq \mathbb R$ and for any $A_s \subset A$, $\int_{A_s} f dt = \int_{A_s} \frac f h dt = 1$ does not hold. Please let me know if I'm making some sense here.