Mass and first moment of the ULE scalar envelope

Put $a(u)=k(-u)$. Apply [F4] with $U=V=\mathbb{R}$, $T(u)=-u$, first to $f(u)=k(u)$ and then to $f(u)=\mathrm{\mid }u\mathrm{\mid }k(u)$. Since $\mathrm{\mid }T(u)\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }{T}^{\mathrm{\prime }}(u)\mathrm{\mid }=1$, $a$ is nonnegative Lebesgue measurable, $\int a=K$ and $\int \mathrm{\mid }u\mathrm{\mid }a(u)du=H$.

Write $\mathcal{L}$ for the Lebesgue sigma-algebra on $\mathbb{R}$. For every Borel target set $B\subseteq \bar{\mathbb{R}}$, the preimages under $(u,v)\mapsto a(u)$ and $(u,v)\mapsto a(v)$ are respectively $a^{-1}(B)\times \mathbb{R}$ and $\mathbb{R}\times a^{-1}(B)$. These are measurable rectangles because $a$ is measurable; [F9] therefore proves that both coordinate pullbacks are $\mathcal{L}\otimes \mathcal{L}$-measurable. Their nonnegative product $p(u,v)=a(u)a(v)$ is measurable by [F10]. Also, $\mathcal{B}(\mathbb{R})\subseteq \mathcal{L}$, so the rectangle-generated definition in [F9] gives $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})\subseteq \mathcal{L}\otimes \mathcal{L}$. By [F11], every Borel subset of ${\mathbb{R}}^2$ is consequently product-measurable. In particular the closed half-plane $\{v\le u\}$, the two open strict half-planes and the closed diagonal $D=\{u=v\}$ are product-measurable. The indicator of each such set $E$ is measurable, since every target preimage of its two-valued indicator is one of $\mathrm{\varnothing },E,E^c,{\mathbb{R}}^2$. Applying [F10] to $p$ and each indicator proves measurability of every restricted integrand used below, and also of ${\mathbf{1}}_D$.

The real line is sigma-finite: the intervals $[-m,m]$ have finite measure and cover it. Apply [F3] to the nonnegative product-measurable function $p(u,v){\mathbf{1}}_{\{v\le u\}}$ from step 2.1. Since $a(u)$ is finite and $\int a=K<\mathrm{\infty }$, [F5] identifies its $v$-section integral with $a(u){\int }_{-\mathrm{\infty }}^{u}a(v)dv$, including when $a(u)=0$. Tonelli makes this marginal measurable and identifies its integral with the product integral, without presupposing integrability of $q$. Multiplication by $4\gamma >0$ preserves measurability by [F10], so $q$ is measurable.

Write $\lambda$ for Lebesgue measure on $\mathbb{R}$ and $\mu =\lambda \otimes \lambda$ for its product measure. Tonelli gives ${\int }_{{\mathbb{R}}^2}a(u)a(v)d\mu (u,v)=K^2$. The diagonal $D=\{u=v\}$ is product-measurable by step 2.1; its sections are singletons, which are null by [F6] under [A1]. Tonelli applied to ${\mathbf{1}}_D$ therefore gives $\mu (D)=0$, and [F7] applied to the nonnegative product-measurable function $a(u)a(v)$ gives ${\int }_{D}a(u)a(v)d\mu =0$. Let $I_{<}$ and $I_{>}$ denote the integrals of $a(u)a(v)$ over $\{v<u\}$ and $\{u<v\}$, respectively. Tonelli applied to $a(u)a(v){\mathbf{1}}_{\{v<u\}}$, exchanging the order of integration and relabelling the two variables, gives $I_{<}=I_{>}$ because the factors commute. The two strict half-planes and $D$ form a measurable partition of ${\mathbb{R}}^2$. Apply [F8] twice to the pointwise decomposition of $a(u)a(v)$ into its restrictions to those three sets: $K^2=I_{<}+I_{>}+0$. Thus $I_{<}=I_{>}=K^2\mathrm{/}2$; applying [F8] once more shows that including $D$ changes neither value. Consequently $\int q=4\gamma K^2\mathrm{/}2=\mathrm{\Gamma }\mathrm{/}2$.

For every $u$, the inner integral in $q(u)$ is at most $\int a=K$. The product $\mathrm{\mid }u\mathrm{\mid }q(u)$ is measurable by [F10]. Hence $0\le q(u)\le 4\gamma Ka(u)$, and monotonicity and positive homogeneity from [F5] give $\int \mathrm{\mid }u\mathrm{\mid }q(u)du\le 4\gamma K\int \mathrm{\mid }u\mathrm{\mid }a(u)du=4\gamma KH=\mathrm{\Gamma }\tau$. The mass from step 4.1 and this first-moment bound are finite.