Let $\sigma: \mathbb R \times \mathbb R \to \mathbb R$ be a Lipschitz continuous function bounded below by some $M > 0$.
Let $W$ be a standard Brownian motion, and let $X$ be the solution to the SDE
$$dX_t = \sigma(t, X_t) dW_t$$
with $X_0 = 0$.
Question: Fix $T > 0$. Does there exist, for every $\varepsilon, h > 0$ a $\delta > 0$ such that
$$\mathbb P\left(\int_{0}^T \mathbf 1_{[-\delta, \delta]} (X_s) ds > h\right) < \varepsilon?$$
I think so. The martingale $X_t$ is a time-changed Brownian motion: $X_t = B_{A_t}$, where $$A_T = \int_0^T \sigma^2(t, X_t) dt$$ and $B_t$ is some Brownian motion. Now write $$ \int_0^T \mathbb 1_{[-\delta,\delta]}(X_t) dt = \int_0^{A_T} \mathbb 1_{[-\delta,\delta]}(B_s) dA^{-1}_s = \int_0^{A_T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{\sigma^2(A^{-1}_s, B_s)} ds \leqslant \int_0^{A_T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{M^2} ds . $$ If $\sigma(t, x)$ is bounded above by some number $N$, then we get $$ \int_0^T \mathbb 1_{[-\delta,\delta]}(X_t) dt \leqslant \int_0^{N^2 T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{M^2} ds , $$ and the desired result follows directly from the corresponding statement for the Brownian motion. The general case requires a more delicate bound on $A_T$. I do not have time now to work out the details, but if need be, I will get back to this when I am more available.
