Levy inversion formula

Let $\mu ={\mathbb{P}}_X$ and define $q(t)={\int }_a^b{e}^{-ity}dy$. Applying FTC to the sine and cosine components gives the stated quotient for $t\mathrm{\ne }0$, while $q(0)=b-a$. The integral expression shows $\mathrm{\mid }q(t)\mathrm{\mid }\le b-a$ and continuity at zero by dominated convergence on $[a,b]$. F7 identifies the compact integrals with Lebesgue integrals; AC covers its assumption and F2.

For $T>0$ the joint integrand $q(t)e^{itx}$ is Borel and its absolute integral against $dt\mu (dx)$ on $[-T,T]\times \mathbb{R}$ is at most $2T(b-a)$. Fubini gives ${\int }_{-T}^{T}q(t){\phi }_X(t)dt={\int }_{\mathbb{R}}(R_T(x-a)-R_T(x-b))\mu (dx),R_T(z)={\int }_{-T}^T\frac{\sin (tz)}{t}dt\mathrm{.}$ Indeed expand the exponentials after multiplication by $e^{itx}$: the imaginary part is an odd function of t and integrates to zero, leaving the two displayed real sine integrals.

F2 bounds the difference of sine kernels uniformly in x and T, and its limit is $\pi (\mathrm{sgn}(x-a)-\mathrm{sgn}(x-b))$. This equals $2\pi$ when $a<x<b$, $\pi$ when $x=a$ or $x=b$, and zero when $x<a$ or $x>b$. Since $\mu$ has mass one, dominated convergence applies to the right-hand side of step 2.1. Division by $2\pi$ proves every term of the stated formula, including the half endpoint atoms.