Vanishing tail control bounds truncated second moments

Zero truncation at a positive level: For a real random variable $X$ and a deterministic level $A>0$, its zero truncation is $X^{(A)}=X{\mathbf{1}}_{\{\mathrm{\mid }X\mathrm{\mid }\le A\}}\mathrm{.}$ The threshold event is measurable because $X$ is measurable and $[-A,A]$ is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus $X^{(A)}$ is a real random variable as in def-random-element-and-real-random-variable. It equals $X$ at both cutoff endpoints and is zero outside the interval. Since $\mathrm{\mid }{X}^{(A)}\mathrm{\mid }\le A$, for every $0<p<\mathrm{\infty }$ its absolute $p$th moment is at most $A^p\mathbb{P}(\mathrm{\Omega })=A^p$. This is not clipping to the endpoints.

For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function: Let $(X,\mathcal{A},\mu )$ be a measure space, let $f:X\to \mathbb{C}$ be measurable, and let $0<p<\mathrm{\infty }$. Then ${\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu =p{\int }_0^{\mathrm{\infty }}{t}^{p-1}\mu (\{\mathrm{\mid }f\mathrm{\mid }>t\})dt=p{\int }_0^{\mathrm{\infty }}{t}^{p-1}{A}_f(t)dt,$ where either side may be $+\mathrm{\infty }$.