[proofplan] We first convert the $p$-th moment of $|X|$ into an integral of its tail probabilities by writing $|X|^p$ as an integral over the sublevel variable $t$ and applying Tonelli's theorem. The assumed Gaussian tail bound then reduces the estimate to a one-dimensional integral involving $t^{p-1}e^{-t^2/a^2}$. A change of variables turns this integral into a Gamma-type integral, and an elementary estimate for that integral gives the required $\sqrt p$ growth. [/proofplan] [step:Rewrite the moment as an integral of tail probabilities] Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,\infty)$ with its Borel $\sigma$-algebra $\mathcal B([0,\infty))$. Define the non-negative [random variable](/page/Random%20Variable) $Y:(\Omega,\mathcal F)\to([0,\infty),\mathcal B([0,\infty)))$ by $Y(\omega):=|X(\omega)|$. For each $\omega\in\Omega$, the one-variable calculus identity gives \begin{align*} Y(\omega)^p=\int_0^{Y(\omega)} p t^{p-1}\,d\mathcal L^1(t) =\int_0^\infty p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathcal L^1(t). \end{align*} Define $H:\Omega\times[0,\infty)\to[0,\infty]$ by $H(\omega,t):=p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}$. The map $H$ is non-negative and measurable with respect to $\mathcal F\otimes\mathcal B([0,\infty))$. Applying Tonelli's theorem for non-negative [measurable functions](/page/Measurable%20Functions) (citing a result not yet in the wiki: Tonelli's theorem), we obtain \begin{align*} \mathbb E[Y^p]=\int_\Omega Y(\omega)^p\,d\mathbb P(\omega). \end{align*} Using the pointwise integral representation of $Y(\omega)^p$ gives \begin{align*} \mathbb E[Y^p]=\int_\Omega\int_0^\infty p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathcal L^1(t)\,d\mathbb P(\omega). \end{align*} Tonelli's theorem changes the order of integration, so \begin{align*} \mathbb E[Y^p]=\int_0^\infty p t^{p-1}\left(\int_\Omega \mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathbb P(\omega)\right)\,d\mathcal L^1(t). \end{align*} Since the inner integral is $\mathbb P(Y\geq t)$, we conclude \begin{align*} \mathbb E[Y^p]=p\int_0^\infty t^{p-1}\mathbb P(Y\geq t)\,d\mathcal L^1(t). \end{align*} [guided] The purpose of this step is to replace the moment $\mathbb E[Y^p]$ by a tail integral, because the hypothesis controls precisely the tail probability $\mathbb P(Y\geq t)$. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $[0,\infty)$ with its Borel $\sigma$-algebra $\mathcal B([0,\infty))$. First define the non-negative random variable $Y:(\Omega,\mathcal F)\to([0,\infty),\mathcal B([0,\infty)))$ by $Y(\omega):=|X(\omega)|$. Since $X$ is measurable and the absolute value map $\mathbb R\to[0,\infty)$ is Borel measurable, $Y$ is measurable. For every fixed $\omega\in\Omega$, ordinary one-variable integration gives \begin{align*} Y(\omega)^p=\int_0^{Y(\omega)} p t^{p-1}\,d\mathcal L^1(t). \end{align*} We rewrite the variable upper limit using an indicator function: \begin{align*} \int_0^{Y(\omega)} p t^{p-1}\,d\mathcal L^1(t) =\int_0^\infty p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathcal L^1(t). \end{align*} Now define $H:\Omega\times[0,\infty)\to[0,\infty]$ by $H(\omega,t):=p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}$. The function $H$ is non-negative and measurable: the factor $t\mapsto p t^{p-1}$ is Borel measurable on $[0,\infty)$, and the set \begin{align*} \{(\omega,t)\in\Omega\times[0,\infty):t\leq Y(\omega)\} \end{align*} is measurable because it is determined by the measurable map $(\omega,t)\mapsto t-Y(\omega)$. Therefore Tonelli's theorem for non-negative measurable functions applies (citing a result not yet in the wiki: Tonelli's theorem). It permits us to exchange the order of integration without first proving integrability. We first write \begin{align*} \mathbb E[Y^p]=\int_\Omega Y(\omega)^p\,d\mathbb P(\omega). \end{align*} Substituting the pointwise representation of $Y(\omega)^p$ gives \begin{align*} \mathbb E[Y^p]=\int_\Omega\int_0^\infty p t^{p-1}\mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathcal L^1(t)\,d\mathbb P(\omega). \end{align*} Tonelli's theorem then gives \begin{align*} \mathbb E[Y^p]=\int_0^\infty p t^{p-1}\left(\int_\Omega \mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathbb P(\omega)\right)\,d\mathcal L^1(t). \end{align*} The inner integral is the probability of the event $\{\omega\in\Omega:Y(\omega)\geq t\}$, so \begin{align*} \int_\Omega \mathbb 1_{\{t\leq Y(\omega)\}}\,d\mathbb P(\omega)=\mathbb P(Y\geq t). \end{align*} Hence \begin{align*} \mathbb E[Y^p] =p\int_0^\infty t^{p-1}\mathbb P(Y\geq t)\,d\mathcal L^1(t). \end{align*} [/guided] [/step] [step:Insert the Gaussian tail bound and reduce to a Gamma integral] Since $Y=|X|$, the hypothesis gives $\mathbb P(Y\geq t)\leq K e^{-t^2/a^2}$ for every $t\geq 0$. Therefore \begin{align*} \mathbb E[Y^p] &\leq pK\int_0^\infty t^{p-1}e^{-t^2/a^2}\,d\mathcal L^1(t). \end{align*} Define the change-of-variables map $\Phi:(0,\infty)\to(0,\infty)$ by $\Phi(t):=t^2/a^2$. With $u=\Phi(t)=t^2/a^2$, we have $t=a\sqrt u$ and \begin{align*} d\mathcal L^1(t)=\frac{a}{2}u^{-1/2}\,d\mathcal L^1(u). \end{align*} Thus \begin{align*} pK\int_0^\infty t^{p-1}e^{-t^2/a^2}\,d\mathcal L^1(t)=pK\int_0^\infty (a\sqrt u)^{p-1}e^{-u}\frac{a}{2}u^{-1/2}\,d\mathcal L^1(u). \end{align*} Collecting powers of $a$ and $u$ gives \begin{align*} pK\int_0^\infty t^{p-1}e^{-t^2/a^2}\,d\mathcal L^1(t)=\frac{pK a^p}{2}\int_0^\infty u^{p/2-1}e^{-u}\,d\mathcal L^1(u). \end{align*} Define the Gamma integral as the map $\Gamma:(0,\infty)\to(0,\infty)$ given by \begin{align*} \Gamma(s):=\int_0^\infty u^{s-1}e^{-u}\,d\mathcal L^1(u). \end{align*} Taking $s=p/2$ gives \begin{align*} \mathbb E[Y^p]\leq K a^p\Gamma\left(\frac p2+1\right), \end{align*} because $\frac p2\Gamma(p/2)=\Gamma(p/2+1)$. This identity follows by [integration by parts](/theorems/210) on $(0,R)$ applied to $u^{p/2}e^{-u}$, then passing to $R\to\infty$: the boundary term at $0$ is $0$, and the boundary term $R^{p/2}e^{-R}$ tends to $0$ as $R\to\infty$. [/step] [step:Bound the Gamma integral by an elementary estimate] Let $r:=p/2$, so $r\geq 1/2$. We claim that \begin{align*} \Gamma(r+1)\leq 3(2r)^r. \end{align*} Indeed, \begin{align*} \Gamma(r+1)=\int_0^\infty u^r e^{-u}\,d\mathcal L^1(u) =\int_0^{2r} u^r e^{-u}\,d\mathcal L^1(u)+\int_{2r}^\infty u^r e^{-u}\,d\mathcal L^1(u). \end{align*} On $(0,2r)$, we have $u^r\leq (2r)^r$ and $e^{-u}\leq 1$, while $\mathcal L^1((0,2r))=2r$. Since $r\geq 1/2$, we have $2r\leq (2r)^r$ when $2r\geq 1$ only if $r\geq 1$; to cover all $r\geq1/2$, use the sharper bound \begin{align*} \int_0^{2r}u^r e^{-u}\,d\mathcal L^1(u)\leq \int_0^{2r}u^r\,d\mathcal L^1(u) =\frac{(2r)^{r+1}}{r+1}\leq 2(2r)^r. \end{align*} On $[2r,\infty)$, the function $u\mapsto u^r e^{-u/2}$ has derivative \begin{align*} \frac{d}{du}\left(u^r e^{-u/2}\right)=u^{r-1}e^{-u/2}\left(r-\frac u2\right), \end{align*} so it is non-increasing for $u\geq 2r$. Hence for $u\geq2r$, \begin{align*} u^r e^{-u}=u^r e^{-u/2}e^{-u/2}\leq (2r)^r e^{-r}e^{-u/2}. \end{align*} Therefore \begin{align*} \int_{2r}^\infty u^r e^{-u}\,d\mathcal L^1(u)\leq (2r)^r e^{-r}\int_{2r}^\infty e^{-u/2}\,d\mathcal L^1(u). \end{align*} Evaluating the exponential integral gives \begin{align*} \int_{2r}^\infty u^r e^{-u}\,d\mathcal L^1(u)\leq 2(2r)^r e^{-2r}. \end{align*} Since $2e^{-2r}\leq 1$ for $r\geq 1/2$, this implies \begin{align*} \int_{2r}^\infty u^r e^{-u}\,d\mathcal L^1(u)\leq (2r)^r. \end{align*} Combining the two estimates gives \begin{align*} \Gamma(r+1)\leq 3(2r)^r. \end{align*} [/step] [step:Take the $p$-th root and conclude the moment bound] Using $r=p/2$, the preceding estimate gives \begin{align*} \Gamma\left(\frac p2+1\right)\leq 3p^{p/2}. \end{align*} Substituting this into the moment estimate yields \begin{align*} \mathbb E[|X|^p]=\mathbb E[Y^p]\leq 3K a^p p^{p/2}. \end{align*} Taking the $p$-th root gives \begin{align*} \left(\mathbb E[|X|^p]\right)^{1/p} \leq 3^{1/p}K^{1/p}a\sqrt p. \end{align*} Since $p\geq1$ and $K\geq1$, we have $3^{1/p}\leq3$ and $K^{1/p}\leq K$. Therefore \begin{align*} \left(\mathbb E[|X|^p]\right)^{1/p} \leq 3K a\sqrt p. \end{align*} This proves the asserted moment growth estimate with the universal numerical constant $C_0=3$ multiplying the factor $K$. [/step]