Please consider if the following pair of inequalities hold $$ g(X) \le \mathbb{P}\left(\xi > X \right) \le h(X) $$ where the domain of integration is over some variable $x$ and $X$ is some constant in this domain.
If $k(x) \ge 0$ is some function will the following hold.
$$ \int_{X}^{\infty}k(x)g'(x)\mbox{d}x \le \int_{X}^{\infty}k(x)f_{\xi}(x)\mbox{d}x \le \int_{X}^{\infty}k(x)h'(x)\mbox{d}x $$ where $g'$ and $h'$ represent derivatives of $g$ and $h$ respectively wrt to $X$ and $f_{\xi}$ represents the probability density function of $\xi$.
I think this should be true but someone suggested that it might not be true.
How can we proof that this hypothesis is true, or alternatively what will be a valid counter example which shows this is false.
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I think i can restate the problem i have, as what are the minimum additional conditions required on $k \ge 0$ such that $\frac{\mbox {d}k}{\mbox {d}x} < 0$ and $k$ is integrable in the domain of integration. So that multiplying it with both sides of an inequality $$ \int_{X}^{\infty}u(x)\mbox{d}x \le \int_{X}^{\infty}v(x)\mbox{d}x $$ preserves the inequality.
I was thinking of collecting the terms $$ 0 \le \int_{X}^{\infty}(v(x) - u(x))\mbox{d}x $$ and the then try use integration by parts after multiplying with $k$ to show the RHS will be satisfy the inequality, but i was not successful, with that.
If these conditions suffice, how can i show that the inequality is preserved. Any suggestions or help are most welcome and appreciated.