Answer by Martin Hairer for On martingale convergence

You can construct a continuous martingale $X$ such that $X(n) = n$ almost surely. Indeed, for $t < 1$, set $Y(t) = B(t/(1-t))$ for $B$ a Brownian motion and then $X(t) = Y(t \wedge \tau)$ with $\tau = \inf\{t>0:Y(t) = 1\}$. Since $Y$ is a continuous martingale and $\tau$ is a stopping time, $X$ is a continuous martingale such that $X(1) = 1$ almost surely. It then suffices to concatenate independent copies of that process.