The SDE for the Brownian bridge is the following:
$$dX_t = \dfrac{b-X_t}{1-t} \, dt+dB_t$$
with the solution
$$X_t = a(1-t)+bt+(1-t)\int_{0}^t \dfrac{dB_s}{1-s}.$$
The expectation and covariance are:
$$\mathbb{E}(X_t) = a+(b-a)t$$
$$\operatorname{Cov}(X_s,X_t) = \min(s,t)-st$$
Now I want to have a look at what happens as $t\rightarrow 1$.
For the expectation and covariance I get
$$\mathbb{E}(X_1) = b,$$
$$\operatorname{Cov}(X_s,X_1) = \min(s,1)-s$$
But I'm having trouble to see what happens with $X_t$. The first two summands clearly go to b, and the last summand should go to 0 as Brownian bridge expression for a Brownian motion suggests. The prove in the last comment using Doob's maximal inequality and Borel-Cantelli is quite short and I don't understand, what's exactly happening there, especially not, where the last equation comes from. Would be great if someone could explain it more exact how I get $$\lim_{t \rightarrow 1} (1-t)\int_0^t \frac{dB_s}{1-s} = 0 \text{ a.s.} $$
Thanks in advance!