Let $W(t),t \in [0,1]$ be a (Standard) Wiener Process. The Brownian Bridge $B(t), t \in [0,1]$ can be constructed via $B(t):=W(t) - t \cdot W(1)$ and is a Gaussian process with zero mean and covariance $\operatorname{ Cov} (B(s),B(t))=\min(s,t)-st$.
Now, let the Integrated Brownian Bridge be defined as
$\overline{B}(t):= \int \limits_{0}^{t}B(s)ds$ for $0 \leq t \leq 1$.
It is stated that $\overline{B}(t), t \in [0,1]$ is a Gaussian process with zero mean and covariance function $$ \operatorname{\overline{B}(s),\overline{B}(t)} = \frac{st \cdot \min(s,t)}{2} - \frac{(\min(s,t)^{3}}{6} - \frac{s^{2}t^{2}}{4},$$ as can be seen here, on page 395.
My Question(s) now: How can I show that the Integrated Brownian Bridge is a Gaussian process, i.e. how do I show the multivariate normality for any $0 \leq t_{1} \leq ... \leq t_{k} \leq 1$, $k \in \mathbb{N}$ give nthe integral expression. Do I have to do it in an "algebraic induction" fashion, i.e. starting with simple functions, than non-negative ones etc. or/and is their a kind of $L^{2}[0,1]$ formulation where I can make use of the Cauchy criterion.
And how to derive the covariance function? As expectations are integrals as well I guess Fubini's theorem is involved, right?
Thanks for any help!