If the sample paths of a martingale are almost surely continuous and not constant on any interval, is it true that they are almost surely not increasing on any interval?
Edit for clarity: Let $(X_t)_{t \geq 0}$ be a continuous martingale that is not constant on any interval. Show that, almost surely, it cannot be strictly increasing on any interval. That is, there exists no $A \in \mathcal{F}$, $\mathbb{P}(A)>0$, such that $$[a,b] \ni t \mapsto X_t(\omega)$$ is strictly increasing for all $\omega \in A$, i.e. $X_t>X_s$ for $t>s\in[a,b]$.
I know that if a martingale is continuous and not constant on any interval, it has unbounded variation over every interval, which implies the result. But it seems that not increasing on any interval is a more natural result of the martingale property, so there should be a more direct way to show this. Is there a way without making reference to the total variation?
All I can conclude so far is in the case where the increments $I_k=X_{1-2^{-k+1}}-X_{1-2^{-k}}$ are independent we have $\mathbb{P}(\forall k,I_k>0)=\prod_k\mathbb{P}(I_k>0)$. If there is a uniform bound $1>B>\mathbb{P}(I_k>0)>0$ then $\mathbb{P}(\forall k,I_k>0)=0$, so $X_t$ is almost surely not strictly increasing over $[0,1]$.