For Brownian motion and any $w, \eta >0$ does there exists $\delta >0$ such that $P\{\exists s,t \in [0,1],|s-t|\le \delta: |B_s-B_t|>\eta \}
For Brownian motion $\{B_t \} $ and any $w, \eta >0$ does there exists $\delta >0$ such that $P\{\exists s,t \in [0,1],|s-t|\le \delta: |B_s-B_t|>\eta \}<w$?
Since Brownian motion has continuous sample paths almost surely, we have that almost surely it has uniformely continuous sample paths on $[0,1] $. This says that almost surley for every $\omega $ we may find an $\delta (\omega) $ such that $|B_t(\omega)-B_s(\omega)|< w $ for $|s-t|\le \delta (\omega) $. Is it possible to find a $\delta $ that works uniformely as per above?
Thanks in advance!
 A: In fact, you can even do much better than this. A simple version of the Kolmogorov continuity theorem says that

Kolmogorov's continuity criterion: Let $X_t$ be an $\mathbb{R}$-valued stochastic process indexed by $t \in [0,1]$. Suppose that for some positive $\alpha, \beta$ and $K$ we have that
  $$\mathbb{E} [|X_t - X_s|^\alpha] \leq K |t-s|^{1 + \beta}$$
  for all $0 \leq s,t \leq 1$. Then $X$ has a modification whose sample paths are almost-surely $\gamma$-Holder continuous for every $\gamma \in (0, \frac{\beta}{\alpha})$.

In the case of Brownian motion, this is usually applied with $\alpha = 4$,$\beta = 1$ and $K = 2$ to see that Brownian motion can be taken to be $\gamma$-Holder continuous for every $\gamma < \frac12$. What is a little less known is that the usual proof of this criterion actually yields bounds on the probability that $\|B\|_{\gamma}$ is large, where $\|\cdot\|_\gamma$ is the $\gamma$-Holder norm on $[0,1]$. 
In the special case of Brownian motion, you get an estimate of the type:
$$\mathbb{P}(\|B\|_\gamma \geq \eta_1) \leq C\eta_1^{-4}$$
for some fixed contant $C$. By an application of Borel-Cantelli, one then even gets that there is a constant $C_1$ such that $\|B\|_\gamma < C_1$ almost surely. 
In particular, we get that 
$$\mathbb{P}(\exists s,t \in [0,1] :|B_t - B_s| > C_1^{\frac{1}{\gamma}} |t-s|^\frac{1}{\gamma}) = 0.$$
Now take $\delta < \eta^\gamma C_1^{-1}$. Then for $|t-s| < \delta$, $C^\frac{1}{\gamma} |t-s|^\frac{1}{\gamma}< \eta$ so that 
\begin{align}
\mathbb{P}(\exists s,t \in [0,1], |s-t|\leq \delta: |B_s-B_t|>\eta ) &\leq  \mathbb{P}(\exists s,t \in [0,1] :|B_t - B_s| > C_1^{\frac{1}{\gamma}} |t-s|^\frac{1}{\gamma}) \\ &= 0
\end{align}
