Let $B$ be one-dimensional Brownian motion on a filtered probability space, and let $s <t$. Intuitively, $B_s$ and $B_t$ should not be independent events, since $B_t - B_s \sim N(0,t-s)$, so the distribution of $B_t$ is going to be centred at the distribution of $B_s$, i.e., it won't be independent.. Is there a better way to rigorously prove this? I can calculate the joint distribution functions as follows:
$$ \begin{aligned} &B_s = x_s, B_t = x_t \iff B_t = x_t, B_s-B_t = x_s-x_t\\ &\implies f_{X_s,X_t} (x_s,x_t) = f_{X_t}(x_t)f_{X_s-X_t}(x_s-x_t) \end{aligned} $$
But I'm not sure if that helps... Please let me know where I should be going and if possible point me to some good resources.
They are correlated: $\operatorname{Cov}(B_s,B_t)=\min \{ s,t \}>0$ for $s,t>0$. How exactly you should prove this depends on your definition of $B_t$; in particular, under one choice of definition, this property is built in. Consider instead the assumption that for $s \leq t$, $B_t-B_s$ is independent of $B_s$ with $N(0,t-s)$ distribution. Then
$$E[B_s B_t]=E[B_s[B_s + (B_t-B_s)]]=E[B_s^2+E[B_s(B_t-B_s)]=s+0=s.$$