Brownian Motion and Continuity Consider a Brownian Motion $(B_{t})_{t\geq0}$. In my lecure notes it says, without proof, that $\mathbb{P}\left(\sup_{t,s\leq N}\left\{ \left|B_{t}-B_{s}\right|:\left|t-s\right|<\delta\right\} <\varepsilon\right)$ converges to $0$ for $\delta$ tending to $0$.
I think it is a consequence of the continuity of the sample paths, but can anyone help with a proof?

Another approach would be to use the continuity of the sample paths directly. As above, I'm assuming you mean that the probability converges to $1$ rather than $0$.
Using the density of the rationals in the reals, and then upwards continuity of the probability measure, we get:
$\lim_{\delta\rightarrow0}\mathbb{P}\left(\sup_{t,s\leq N}\left\{ \left|B_{t}-B_{s}\right|:\left|t-s\right|<\delta\right\} <\varepsilon\right) = \lim_{\delta\rightarrow0}\mathbb{P}\left(\bigcap_{t,s\in\left[0,N\right]}\left\{ \left|B_{t}-B_{s}\right|<\varepsilon:\left|t-s\right|<\delta\right\} \right)
 = \lim_{n\rightarrow\infty}\mathbb{P}\left(\bigcap_{t,s\in\left[0,N\right]\cap\mathbb{Q}}\left\{ \left|B_{t}-B_{s}\right|<\varepsilon:\left|t-s\right|<\frac{1}{n}\right\} \right)
 = \mathbb{P}\left(\bigcup_{n\geq0}\bigcap_{t,s\in\left[0,N\right]\cap\mathbb{Q}}\left\{ \left|B_{t}-B_{s}\right|<\varepsilon:\left|t-s\right|<\frac{1}{n}\right\} \right)
 = \mathbb{P}\left(\exists\delta>0\;\mbox{s.t.}\;\left|B_{t}-B_{s}\right|<\varepsilon\;\mbox{whenever}\;\left|t-s\right|<\delta\;\mbox{and}\; t,s\in[0,N]\right)$
The latter probability is $1$
  since $t\mapsto B_{t}\left(\omega\right)$
  is uniformly continuous on $[0,N]$.
