Lemma 1.1.8.label Let $\lambda \in \sps^{*}$, then there exists $h_{\lambda} \in \ccm$ such that $\angles{\theta, \lambda}_{\sps} = \angles{\theta, h_\lambda}_{\ccm}$ for all $\theta \in \ccm$, which is given by
\[h_{\lambda}(t) = \int_{(0,t]}\lambda((s, \infty))ds\]
with
\[\norm{h_\lambda}_{\ccm}^{2} = \iint_{[0, \infty)}(s \wedge t)d\lambda(s)d\lambda(t) = \ev^{\wien}(\abs{\lambda}^{2})\]
thus
\[\wh \wien(\lambda) = \exp\braks{-\frac{\norm{h_\lambda}_{\ccm}}{2}^2}\]
Moreover, for each $\theta \in \sps$,
\[\angles{\theta, \lambda}_{\sps} = \int_{0}^{\infty} Dh_{\lambda} d\theta\]
is an improper Riemann-Stieltjes integral.
Proof. Firstly, $\lambda$ is bounded, so $h_{\lambda}$ is absolutely continuous and $Dh_{\lambda}(s) = \lambda((s, \infty))$ by the Lebesgue differentiation theorem. From here,
\begin{align*}\norm{h_\lambda}_{\ccm}^{2}&= \int_{[0, \infty)}\abs{\lambda((t, \infty))}^{2}dt = \int_{[0, \infty)}\iint_{[t, \infty)^2}d\lambda(r)d\lambda(s)dt \\&= \iiint_{[0, \infty)^3}\one_{\bracs{r, s \ge t}}(r, s, t)d\lambda(r)d\lambda(s)dt \\&= \iint_{[0, \infty)^2}\int_{[0, \infty)}\one_{\bracs{t \le r \wedge s}}(r, s, t)dtd\lambda(r)d\lambda(s) \\&= \iint_{[0, \infty)^2}s \wedge t d\lambda(r)d\lambda(s) = \ev^{\wien}[\abs{\lambda}^{2}]\end{align*}
so $h_{\lambda} \in \ccm$ and
\begin{align*}\angles{h, \lambda}_{\sps}&= \int_{[0, \infty)}h d\lambda = \int_{[0, \infty)}\int_{[0, t]}Dh(s)dsd\lambda(t) \\&= \int_{[0, \infty)}\int_{[0, \infty)}\one_{\bracs{s \le t}}(s, t)Dh(s)d\lambda(t) ds \\&= \int_{[0, \infty)}Dh(s)\lambda((s, \infty))ds = \angles{\theta, h_\lambda}_{\ccm}\end{align*}
Now let $F(t) = \lambda((0, t])$, then $Dh_{\lambda}(t) = \lambda((0, \infty)) - F(t)$ and
\[\int_{0}^{\infty} \theta(t) d\lambda(t) = \int_{0}^{\infty} \theta(t)dF(t) = -\int_{0}^{\infty} \theta(t) dDh_{\lambda}(t)\]
can be expressed as improper Riemann-Stieltjes integrals. Using the by parts formula for Riemann-Stieltjes integrals,
\[\int_{0}^{T} \theta(t) dDh_{\lambda}(t) = \theta(T)Dh_{\lambda}(T) - \int_{0}^{T}Dh_{\lambda}(t)d\theta(t)\]
where since $\theta(T)/T \to 0$ as $T \to \infty$ and $(1 + t) \in L^{1}(\abs{\lambda})$,
\[\limv{T}\abs{\theta(T)}\abs{\lambda}((T, \infty)) \le \limv{T}\frac{\abs{\theta(T)}}{1 + T}\int_{(0, \infty)}(1 + t)d\abs{\lambda}(t) = 0\]
Therefore
\[\angles{h, \lambda}_{\sps} = -\int_{0}^{\infty} \theta(t)dDh_{\lambda}(t) = \int_{0}^{\infty} Dh_{\lambda}(t)d\theta(t)\]
$\square$