[proofplan] We express the moment $\mathbb E[X^p]$ as an integral of the tail probabilities $\mathbb P(X>t)$. The assumed bound is then used in two different regimes: for $0<t\le A$ the probability is bounded by $1$, while for $t>A$ the polynomial decay gives an integrable power because $p<q$. Adding the two contributions gives the stated constant. [/proofplan]

[step:Represent the moment as a tail integral]Fix $p\in(0,q)$. Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the Borel $\sigma$-algebra $\mathcal B((0,\infty))$. Define the nonnegative [random variable](/page/Random%20Variable) $Y:\Omega\to[0,\infty]$ by \begin{align*} Y(\omega)=X(\omega)^p\quad\text{for every }\omega\in\Omega. \end{align*} For each $\omega\in\Omega$, \begin{align*} Y(\omega)=\int_0^\infty \mathbb{1}_{(0,Y(\omega))}(s)\,d\mathcal L^1(s). \end{align*} Applying Tonelli's theorem to the nonnegative measurable map $(\omega,s)\mapsto \mathbb{1}_{(0,Y(\omega))}(s)$ on the product [measure space](/page/Measure%20Space) $(\Omega\times(0,\infty),\mathcal F\otimes\mathcal B((0,\infty)),\mathbb P\otimes\mathcal L^1)$ gives \begin{align*} \mathbb E[X^p]=\int_\Omega Y(\omega)\,d\mathbb P(\omega). \end{align*} The same application of Tonelli's theorem yields \begin{align*} \int_\Omega Y(\omega)\,d\mathbb P(\omega)=\int_0^\infty \mathbb P(Y>s)\,d\mathcal L^1(s). \end{align*} Since $Y=X^p$ and the map $t\mapsto t^p$ is strictly increasing on $[0,\infty]$, we have $\{Y>s\}=\{X>s^{1/p}\}$ for $s>0$. Using the one-dimensional change-of-variables theorem for Lebesgue integrals in its nonnegative measurable, possibly extended-valued, form with the $C^1$ bijection $\varphi:(0,\infty)\to(0,\infty)$ defined by $\varphi(t)=t^p$, whose derivative is $\varphi'(t)=p t^{p-1}$, we obtain \begin{align*} \mathbb E[X^p] =p\int_0^\infty t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t). \end{align*}[/step]

[guided]The goal is to turn information about the tail probabilities $\mathbb P(X>t)$ into information about the moment $\mathbb E[X^p]$. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on the Borel $\sigma$-algebra $\mathcal B((0,\infty))$. To do this, define the map $Y:\Omega\to[0,\infty]$ by \begin{align*} Y(\omega)=X(\omega)^p\quad\text{for every }\omega\in\Omega. \end{align*} This is a nonnegative measurable random variable because $X$ is measurable and $r\mapsto r^p$ is Borel measurable on $[0,\infty]$. For a fixed $\omega\in\Omega$, the number $Y(\omega)$ is the length of the interval $(0,Y(\omega))$. Therefore \begin{align*} Y(\omega)=\int_0^\infty \mathbb{1}_{(0,Y(\omega))}(s)\,d\mathcal L^1(s). \end{align*} The integrand $(\omega,s)\mapsto \mathbb{1}_{(0,Y(\omega))}(s)$ is nonnegative and measurable on the product measure space $(\Omega\times(0,\infty),\mathcal F\otimes\mathcal B((0,\infty)),\mathbb P\otimes\mathcal L^1)$, so Tonelli's theorem applies. Hence \begin{align*} \mathbb E[X^p]=\int_\Omega Y(\omega)\,d\mathbb P(\omega). \end{align*} Substituting the interval-length representation of $Y(\omega)$ gives \begin{align*} \int_\Omega Y(\omega)\,d\mathbb P(\omega)=\int_\Omega\int_0^\infty \mathbb{1}_{(0,Y(\omega))}(s)\,d\mathcal L^1(s)\,d\mathbb P(\omega). \end{align*} Tonelli's theorem permits interchange of the two nonnegative integrals, so \begin{align*} \int_\Omega\int_0^\infty \mathbb{1}_{(0,Y(\omega))}(s)\,d\mathcal L^1(s)\,d\mathbb P(\omega)=\int_0^\infty\int_\Omega \mathbb{1}_{(s,\infty)}(Y(\omega))\,d\mathbb P(\omega)\,d\mathcal L^1(s). \end{align*} By the definition of probability of the event $\{Y>s\}$, \begin{align*} \int_0^\infty\int_\Omega \mathbb{1}_{(s,\infty)}(Y(\omega))\,d\mathbb P(\omega)\,d\mathcal L^1(s)=\int_0^\infty \mathbb P(Y>s)\,d\mathcal L^1(s). \end{align*} Since $Y=X^p$ and $r\mapsto r^p$ is strictly increasing on $[0,\infty]$, the event $\{Y>s\}$ is exactly the event $\{X>s^{1/p}\}$ for every $s>0$. Thus \begin{align*} \mathbb E[X^p]=\int_0^\infty \mathbb P(X>s^{1/p})\,d\mathcal L^1(s). \end{align*} Now apply the one-dimensional change-of-variables theorem for Lebesgue integrals in its nonnegative measurable, possibly extended-valued, form with the map $\varphi:(0,\infty)\to(0,\infty)$ defined by $\varphi(t)=t^p$. This map is a $C^1$ bijection, and its derivative is $\varphi'(t)=p t^{p-1}$. The integrand $s\mapsto \mathbb P(X>s^{1/p})$ is nonnegative and measurable, so this form of the theorem applies. Therefore the transformed measure factor is $p t^{p-1}\,d\mathcal L^1(t)$, and \begin{align*} \mathbb E[X^p] =p\int_0^\infty t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t). \end{align*} This identity is the bridge between the tail estimate and the desired moment estimate.[/guided]

[step:Split the tail integral at the scale $A$] Using the tail representation and splitting the integration domain $(0,\infty)=(0,A]\cup(A,\infty)$ gives \begin{align*} \mathbb E[X^p] &=p\int_0^A t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t) +p\int_A^\infty t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t). \end{align*} On $(0,A]$, the probability bound $\mathbb P(X>t)\le 1$ gives \begin{align*} p\int_0^A t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t) \le p\int_0^A t^{p-1}\,d\mathcal L^1(t) =A^p. \end{align*} On $(A,\infty)$, the assumed polynomial tail bound gives \begin{align*} \mathbb P(X>t)\le \left(\frac{A}{t}\right)^q=A^q t^{-q}. \end{align*} Therefore \begin{align*} p\int_A^\infty t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t) &\le pA^q\int_A^\infty t^{p-q-1}\,d\mathcal L^1(t). \end{align*} Since $p<q$, the exponent $p-q-1$ is strictly less than $-1$, and \begin{align*} \int_A^\infty t^{p-q-1}\,d\mathcal L^1(t) =\frac{A^{p-q}}{q-p}. \end{align*} Thus \begin{align*} p\int_A^\infty t^{p-1}\mathbb P(X>t)\,d\mathcal L^1(t) \le \frac{p}{q-p}A^p. \end{align*} [/step]

[step:Add the two bounds to obtain the moment estimate] Combining the estimates on $(0,A]$ and $(A,\infty)$ yields \begin{align*} \mathbb E[X^p]\le A^p+\frac{p}{q-p}A^p. \end{align*} Factoring out $A^p$ gives \begin{align*} A^p+\frac{p}{q-p}A^p=A^p\left(1+\frac{p}{q-p}\right). \end{align*} This is the desired estimate for the fixed $p\in(0,q)$. Since $p$ was arbitrary in $(0,q)$, the conclusion holds for every $p\in(0,q)$. [/step]