Let $\{W_t\}$ be standard Brownian motion. Let $B_t=W_t-tW_1$ be Brownian bridge on $[0,1]$. Let $\mu$ be a Borel probability measure on $[0,1]$. I want to show that $\int_0^1 B_t\mu(dt)$ is a Gaussian random variable.
Motivation
If $\mu$ is a finite linear combination of point masses $\mu=\sum\alpha_i\delta_{t_i}$ for $0\le t_1<\cdots<t_n\le1$, the result is clear because the integral can be written as a linear combination of the independent Gaussian random variables $W_0,W_{t_1}-W_0,\cdots,W_1-W_{t_n}$. I'm wondering whether this generalizes.
First attempt
Write
$$\begin{align}
\int_0^1B_t\mu(dt)
&=\int_0^1W_t-tW_1\mu(dt)\\
&=\int_0^1W_t\,\mu(dt)-W_1\int_0^1 t\mu(dt)\\
&=\int_0^1\left(\int_0^tdW_s\right)\mu(dt)-W_1\int_0^1 t\mu(dt)
\end{align}$$
and then use Stochastic Fubini's theorem to interchange the two integrals. This shows $\int_0^1 B_t\mu(dt)$ is a difference of two Gaussians, but that need not be a Gaussian in general.
Second attempt
Write
$$\begin{align}
\int_0^1B_t\mu(dt)
&=\int_0^1 \left(\int_0^1\frac{1-t}{1-s}\mathbf1_{[0,t]}(s)d W_s\right)\mu(d t)
\end{align}$$
and use Stochastic Fubini's theorem to interchange the intergrals.
Problem
I'm not sure the theorem applies in the second case. The statement of it in my textbook is the following:
Exercise (Fubini's theorem for stochastic integrals): Let $(U,\mathcal U,\mu)$ be a finite measure space, and $f(t,u)$ a jointly measurable function on $\mathbb R_+\times U$. Prove the statement
$$\begin{align}
\int_U\left(\int_0^T f(s,u) d W_s\right)\mu(d u)
=\int_0^T \left(\int_U f(s,u)\mu( d u)\right) d W_s
\end{align}$$
under some natural assumptions.
Could anyone point me towards either a proof that Fubini holds in the second attempt, or another proof altogether that $\int_0^1 B_t\mu(dt)$ is Gaussian?
Best Answer
This is equivalent to saying that a Brownian bridge is a Gaussian process on the space $C[0,1]$, the dual space of which is the collection of all Borel regular signed measures.
Anyway, let $\mu$ be a Borel probability measure on $[0,1]$. It can be approximated by a convex combination of Dirac deltas, in weak convergence of measures. An approximating sequence $\mu_n$ converging weakly to the measure $\mu$ implies that the distribution of $\int_0^1 B_t\mu_n(dt)$ converges to the distribution of $\int_0^1 B_t\mu(dt)$. From Limit of Gaussian random variables is Gaussian? if we can show that $\int_0^1 B_t\mu_n(dt)$ is Gaussian then the limit is Gaussian. Note that $\int_0^1 B_t\mu_n(dt)$ is just a linear combination of $B_t$, which is Gaussian because $B$ is a Gaussian process.