Markov semigroup for normal distributed kernels Let $\alpha,\sigma^2>0$. I want to show that the kernels defined by
$$K_t(x,\cdot):=\mathcal{N}\big(xe^{-\alpha t},\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})\big)\quad\text{for}\quad t>0$$
$$K_0(x,\cdot)=\varepsilon_x$$
form a Markov semigroup, i.e. for all $(x,B)\in\mathbb{R}\times\mathcal{B}(\mathbb{R})$ and for all $s,t\in \mathbb{R}_+$ it holds
$$K_{s+t}(x,B)=\int_\mathbb{R}K_t(y,B)K_s(x,dy)$$
Clearly we have
$$K_{0+t}(x,B)=\int_\mathbb{R}K_t(y,B) \varepsilon_{x}(dy)=K_t(x,B)$$
and also
$$K_{s+0}(x,B)=\int_\mathbb{R}\varepsilon_y(B)K_s(x,dy)=K_s(x,B)$$
For both $s,t>0$ my attempt is following:
\begin{align}\int_\mathbb{R}K_t(y,B)K_s(x,dy)&=\frac{1}{\frac{\pi\sigma^2}{\alpha}\sqrt{1-e^{-2at}-e^{-2as}+e^{-2a(t+s)}}}\\
&\quad\cdot\int_\mathbb{R}\bigg(\int_{B-ye^{-\alpha t}}\exp\bigg(\frac{z^2}{\frac{\sigma^2}{\alpha}(1-e^{-\alpha t})}\bigg)dz\bigg)\exp\bigg(\frac{(y-xe^{-\alpha s})^2}{\frac{\sigma^2}{\alpha}(1-e^{-\alpha s})}\bigg)dy
\end{align}
but I do not know where I have to go from here... I am grateful for any advice or help. Thanks in advance!
 A: Addendum. If we just want to see it's true, then of course it's true because it's the transition pdf of an OU process. I thought the purpose of the exercise was to "double check" it by direct integration.

We want to show (by direct integration)
$$
\int_{\mathbb R}K_t(y,B)K_s(x,dy)=K_{s+t}(x,B).
$$
Note that
\begin{align}
\int_{\mathbb R}K_t(y,B)K_s(x,dy)
& = 
\int_{\mathbb R}\int_B
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}}
e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}}
dz
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}}
e^{- \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}}
dy\\
& =
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}}
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}}
\int_B
\int_{\mathbb R}
e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}
- \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}
}
dy dz
\end{align}
where in the second equality I changed the order of integration. And
from this point on, it's mechanical. First off, the exponent can be rewritten as
$$
- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}
- \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}
= 
- \frac{1-e^{-2\alpha(s+t)}}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})(1-e^{-2\alpha s})}
\left(
y-\frac{ze^{-\alpha}(1-e^{-2\alpha s}) + xe^{-\alpha s}(1-e^{-2\alpha t})}{1-e^{-2\alpha(s+t)}}
\right)^2
- \frac{(z-xe^{-\alpha(s+t)})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}.
$$
So
\begin{align}
\int_{\mathbb R}K_t(y,B)K_s(x,dy)
& =
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}}
\int_B
\frac{\sqrt{1-e^{-2\alpha (s+t)}}}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})(1-e^{-2\alpha s})}}
%%
\int_{\mathbb R}
e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}
- \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}
}
dy dz\\
& = 
\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}}
\int_B
e^{- \frac{(z-xe^{-\alpha(s+t)})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}}
dz\\
& = K_{s+t}(x,B).
\end{align}
A: The notation makes everything look much more grim than it actually is, so lets quickly define
$$
m(t, x) = e^{-at}x, \qquad v(t) = \frac{\sigma^2}{2a}\left(1-e^{-2at}\right).
$$
Then if you form the double integral, switch the order of integration and then inspect the inner integral
$$
\int_{\mathbb{R}}\mathcal{N}(z \mid m(t, y), v(t))\mathcal{N}(y \mid m(s, x) , v(s) )\mathrm{d}y,
$$
you see this is exactly the same as what we would have in the linear Gaussian model
$$
y \sim \mathcal{N}(y \mid \mu_0, \sigma^2_0), \qquad z\mid y \sim \mathcal{N}(z \mid Ay + b, \sigma_1^2),
$$
and in particular for the linear Gaussian model you have the marginal distribution
$$
p(z) = \mathcal{N}(z \mid A\mu_0 + b, \sigma_1 + A^2\sigma_0).
$$ 
Applying that in this case you have
$$
\begin{align}
p(z) &= \mathcal{N}(z \mid e^{-at}m(s, x), v(t) + e^{-2at}v(s)) \\
\end{align}
$$
then you have
$$
e^{-at}m(s, x) = e^{-at}e^{-as}x =e^{-a(t+s)}x
$$
and
$$
\begin{align}
v(t) + e^{-2at}v(s) &=\frac{\sigma^2}{2a}\left( 1-e^{-2at}\right) + e^{-2at}\frac{\sigma^2}{2a}(1-e^{-2as}) \\
&=\frac{\sigma^2}{2a}\left(1 - e^{-2at} + e^{-2at}-e^{-2a(t+s)} \right) \\
&= \frac{\sigma^2}{2a}\left(1-e^{-2(a+s)}\right).
\end{align}
$$
