I have a question about the Laplace transform of a measure and a semigroup.
Let $(\eta_{t})_{t>0}$ be a family of measures on $(]0,\infty[,\mathcal{B}(]0,\infty[))$.
The Laplace transform $\mathcal{L}\eta_{t}(\lambda)$ of $\eta_{t}$ is the real valued function defined by
\begin{eqnarray*} \mathcal{L}\eta_{t}(\lambda)=\int_{0}^{\infty}\exp(-\lambda s)\eta_{t}(ds),\quad \lambda>0\cdots(\star) \end{eqnarray*}
We assume that $\mathcal{L}\eta_{t}(\lambda)=\exp[-t\lambda]\cdots(\star \star)$.
Let $(p_{t})_{t>0}$ be the Brownian semigroup on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, i.e. , for $f$ (Bounded, Borel function)
\begin{eqnarray*} p_{t}f(x)=\int_{\mathbb{R}} f(y)(2 \pi t)^{-1/2} \exp \left[\frac{-|x-y|^{2}}{2t} \right]dy,\quad x\in\mathbb{R} \end{eqnarray*}
I want to show the following equation: \begin{eqnarray*} p_{t}f=\int_{0}^{\infty}p_{s}f \,\eta_{t}(ds) \end{eqnarray*}
My process: \begin{eqnarray*} \int_{0}^{\infty}p_{s}f \,\eta_{t}(ds)&=&\int_{0}^{\infty} \int_{\mathbb{R}} f(y)(2 \pi s)^{-1/2} \exp \left[\frac{-|x-y|^{2}}{2s} \right]dy\, \eta_{t}(ds)\\ &=&\int_{\mathbb{R}}f(y) \left[ \int_{0}^{\infty} (2 \pi s)^{-1/2} \exp \left[\frac{-|x-y|^{2}}{2s} \right]\, \eta_{t}(ds) \right]dy\\ \end{eqnarray*}
Can I rewrite the following integration in the form of $(*)$?
\begin{eqnarray*} \int_{0}^{\infty} (2 \pi s)^{-1/2} \exp \left[\frac{-|x-y|^{2}}{2s} \right]\, \eta_{t}(ds) \end{eqnarray*}