I am given the process $X_t = B_t -\int_0^t \frac{B_u}{u}du$
How can I show that it is gaussian, given a standard continuos Brownian motion $B$?
As I know that $sB_{1/s} \rightarrow 0$ as $ s \rightarrow 0^+ $almost surely, I can conclude that the integrand is continuos a.s. (by change of variables?) and therefore I'd write it as the a.s. limit of Riemann sums from which I'd conclude gaussianity. Is this correct?
As a start (acknowledging that the integral is improper — the rough order of magnitude of $B_u$ for small $u$ is $\sqrt{u}$, so $\lim_{u\to 0}B_u/u$ doesn't exist) show that $I_\epsilon:=\int_\epsilon^t {B_u\over u}\,ds$ is Gaussian (even jointly with $B_t$) by using a Reimann sum approximation. As a by-product you should be able to compute the variance of $I_\epsilon$ and the covariance of $I_\epsilon$ with $B_t$, to conclude that the Gaussian pair $(B_t,I_\epsilon)$ converges to $(B_t,I)$ which is bivariate normal with zero means and covariance matrix $\left[\matrix{t&t\cr t&2t\cr}\right]$. From this you see that $X_t$ is Gaussian with mean 0 and variance $t$. Repeat the argument to find the joint Gaussian distribution of $(X_{t_1},X_{t_2},\ldots,X_{t_n})$ for $0<t_1<t_2<\cdots<t_n$.