Let $X\ge0$ be a random variable with distribution function $F(t)$ such that $F(t)<1$ for all $t\in\mathbb{R}$ and, for some $\eta\in (1,\infty)$, \begin{align*} \lim_{t\to\infty}\frac{1-F(\eta t)}{1-F(t)}=0. \end{align*} Show that $E[X^m]<\infty$ for any $m\in(0,\infty)$.
I have shown that $EX<\infty$, as done below:
Note that our assumption is that: $\lim\limits_{t\to\infty}\frac{P(X>\eta t)}{P(X>t)}\stackrel{(*)}{=}0$, we first show that $(*)\implies E[X]<\infty$. By $(*)$ $\exists$ an $s\in\mathbb{N}$ such that: \begin{align*} \frac{P(X>\eta t)}{P(X>t)}<\frac{1}{2\eta}\,\,\text{for all $t\ge s$} \end{align*} Now note that: \begin{align*} \int_{s\eta^n}^{s\eta^{n+1}}P(X>t)\,dt&\le P(X>s\eta^n)(s\eta^{n+1}-s\eta^{n})\\ &=s\eta^n(\eta-1)P(X>s\eta^n)\\ &=s\eta^n(\eta-1)\frac{P(X>s\eta^{n})}{P(X>s\eta^{n-1})}\frac{P(X>s\eta^{n-1})}{P(X>s\eta^{n-2})}...\frac{P(X>s\eta)}{P(X>s)}P(X>s)\\ &\le s\eta^n(\eta-1)\frac{1}{2\eta}\frac{1}{2\eta}...\frac{1}{2\eta}P(X>s)\,\,\text{since:}\\ &\text{$\bigg|\frac{s\eta^{n-k}}{s\eta^{n-k-1}}\bigg|=\eta$ and $s\eta^{n-k}=\eta(s\eta^{n-k-1})$ where $s\eta^{n-k-1}\ge s$ as $\eta>1$}\\ &\le s(\eta-1)\eta^n\frac{1}{(2\eta)^n}\quad\text{as $P(X>s)\le1$}\\ &=\frac{s(\eta-1)}{2^n}\\ \end{align*} Thus, \begin{align*} &\int_{s\eta}^{\infty}P(X>t)\,dt=\sum_{n\ge1}\int_{s\eta^n}^{s\eta^{n+1}}P(X>t)\,dt\le s(\eta-1)\sum_{n\ge1}\frac{1}{2^n}=s(\eta-1)<\infty \end{align*} Hence, \begin{align*} EX=\int_{0}^{\infty}P(X>t)\,dt=\int_{0}^{s\eta}P(X>t)\,dt+\int_{s\eta}^{\infty}P(X>t)\,dt\le s\eta+s(\eta-1)<\infty\,\,\text{as we wished to show}. \end{align*}
However, I cannot figure out how to extend this result to $EX^m<\infty$, any help here would be greatly appreciated. My thoughts on an extension are as follows, if we can show that \begin{align*} \lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}=0 \end{align*} Then replacing $X$ with $X^m$ in our above argument finishes the proof, but I cannot show that this limit is zero. Here is what I have \begin{align*} \lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}&=\lim_{t\to\infty}\frac{P(X>(\eta t)^{1/m})}{P(X>t^{1/m})}\\ &=\lim_{t\to\infty}\frac{P(X>{\eta}^{1/m} t^{1/m})}{P(X>t^{1/m})}\\ &=\lim_{z\to\infty}\frac{P(X>{\eta}^{1/m}\cdot z)}{P(X>z)}\quad\text{since $z=t^{1/m}\to\infty$ as $t\to\infty$} \end{align*} But now $\eta>1$ implies that $\eta^{1/m}<\eta$ and so \begin{align} X>\eta z\implies X>\eta^{1/m}z \end{align} And so $P(X>\eta z)\le P(X>\eta^{1/m}z)$, hence \begin{align*} \lim_{t\to\infty}\frac{P(X^m>\eta t)}{P(X^m>t)}=\lim_{z\to\infty}\frac{P(X>{\eta}^{1/m}\cdot z)}{P(X>z)}\ge \lim_{z\to\infty}\frac{P(X>\eta z)}{P(X>z)}=0 \end{align*} and so the inequality is going to the wrong way.
Only slight modifications will be needed for the general case $m\in(0,\infty)$.
First, a change of variables $u\equiv t^{1/m}$ implies that \begin{align*} \mathbb E[X^m]=\int_0^{\infty}\mathbb P[X^m>t]\,\mathrm dt=\int_0^{\infty}\mathbb P[X>t^{1/m}]\,\mathrm dt=\int_0^{\infty}\mathbb P[X>u]m u^{m-1}\,\mathrm du. \end{align*} Second, take $s>0$ so large that \begin{align*} \frac{\mathbb P[X>\eta t]}{\mathbb P[X>t]}<\frac{1}{2\eta^m}\quad\text{for all $t\geq s$.} \end{align*} Third, for any $n\in\{0,1,2,\ldots\}$, \begin{align*} \int_{s\eta^n}^{s\eta^{n+1}}\mathbb P[X>u]m u^{m-1}\,\mathrm du&\leq\mathbb P[X>s\eta^n]\int_{s\eta^n}^{s\eta^{n+1}}m u^{m-1}\,\mathrm du\\ &=\mathbb P[X>s\eta^n]\left[(s\eta^{n+1})^m-(s\eta^{n})^m\right]\\ &=\mathbb P[X>s\eta^n]s^m\eta^{nm}(\eta^m-1)\\ &\leq\frac{\mathbb P[X>s]}{2^n\eta^{nm}}s^m\eta^{nm}(\eta^m-1)\\ &=\mathbb P[X>s]\frac{s^m(\eta^m-1)}{2^n}\\ &\leq\frac{s^m(\eta^m-1)}{2^n}. \end{align*} Finally, do the summation as in the $m=1$ case.