Proposition 19.6.19.F397AFA38927 Let $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. Let $m\in \mathbb {Z}_{>0}$. Then the following extended moment identity holds:
FCEE1C9E921F
19.6.20
\begin{equation} \int _{\mathbb {R}}x^ m \operatorname{DH}(\Delta _{\bullet })(x)=\Delta _{\max }^ m\operatorname{vol}\Delta +m\int _{-\infty }^{\Delta _{\max }} \tau ^{m-1}(\operatorname{vol}\Delta _{\tau }-\operatorname{vol}\Delta )\, \mathrm{d}\tau \end{equation}
and
842F6248A5C4
19.6.21
\begin{equation} \int _{\mathbb {R}}\operatorname{DH}(\Delta _{\bullet })=\operatorname{vol}\Delta . \end{equation}
Proof.
In fact, (19.6.21) follows immediately from the definition. Let $G=G[\Delta _{\bullet }]$. We first assume that $G$ is bounded from below. Then (19.6.20) follows from a straightforward computation:
\[ \begin{aligned} & \int _{\mathbb {R}}x^ m \operatorname{DH}(\Delta _{\bullet })(x)\\ =& \int _{\Delta } G[\Delta _{\bullet }](a)^ m\, \mathrm{d}\operatorname{vol}(a)\\ =& \int _{\Delta } \left( \Delta _{\max }^ m-\int _{G[\Delta _{\bullet }](a)}^{\Delta _{\max }} m\tau ^{m-1}\, \mathrm{d}\tau \right)\, \mathrm{d}\operatorname{vol}(a)\\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{\mathbb {R}}\int _{\Delta } \mathds {1}_{[G[\Delta _{\bullet }](a),\Delta _{\max }]}(\tau )\tau ^{m-1}\, \mathrm{d}\operatorname{vol}(a)\, \mathrm{d}\tau \\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{-\infty }^{\Delta _{\max }}\int _{\Delta \setminus \Delta _{\tau }} \tau ^{m-1}\, \mathrm{d}\operatorname{vol}(a)\, \mathrm{d}\tau \\ =& \Delta _{\max }^ m\operatorname{vol}\Delta -m\int _{-\infty }^{\Delta _{\max }} \tau ^{m-1}\left(\operatorname{vol}\Delta -\operatorname{vol}\Delta _{\tau }\right)\, \mathrm{d}\tau . \end{aligned} \]
In general, set $G_ A=G\lor (-A)$. The same pointwise identity applied to $G_ A$ gives
\[ \int _{\Delta }G_ A^ m\, \mathrm{d}\operatorname{vol}=\Delta _{\max }^ m\operatorname{vol}\Delta +m\int _{-A}^{\Delta _{\max }}\tau ^{m-1}(\operatorname{vol}\Delta _{\tau }-\operatorname{vol}\Delta )\, \mathrm{d}\tau . \]
Letting $A\to \infty $ gives (19.6.20): if $m$ is even, this follows from monotone convergence applied to $G_ A^ m\nearrow G^ m$; if $m$ is odd, apply monotone convergence to $\Delta _{\max }^ m-G_ A^ m\nearrow \Delta _{\max }^ m-G^ m$. Thus both sides of (19.6.20) are interpreted in the corresponding extended sense.
$\square$
Pome
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