How can we show that the sample paths of standard Brownian motion are continuous? I know the path of Brownian mothion is continuous in probability if and only if ,for every $\delta>0$ and $t>0$,
$$\lim_{\Delta t\to 0}P(|B(t+\Delta t)-B(t)|\ge \delta)=0$$
I  can't continue it . I have seen some proofs of this theorem in this site but I want to prove it by definition.
So thanks
 A: Since $\frac{B(t+\Delta t) -B(t)}{\sqrt{\Delta t}} \sim N(0, 1)$, then 
\begin{align*}
&\ P(|B(t+\Delta t) -B(t)|\ge \delta) \\
=&\ P\left(\frac{B(t+\Delta t) -B(t)}{\sqrt{\Delta t}}\ge \frac{\delta}{\sqrt{\Delta t}}\right) + P\left(\frac{B(t+\Delta t) -B(t)}{\sqrt{\Delta t}}\le \frac{-\delta}{\sqrt{\Delta t}}\right)\\
=&\ 1-\Phi\left(\frac{\delta }{\sqrt{\Delta t}}\right) + \Phi\left(\frac{-\delta }{\sqrt{\Delta t}}\right),
\end{align*}
where $\Phi$ is the cumulative distribution function of a standard normal random vaiable.
It is now obvious that
\begin{align*}
\lim_{\Delta t \rightarrow 0}P(|B(t+\Delta t) -B(t)|\ge \delta) = 0.
\end{align*}
A: Other way
We know $B_t$ is martingale thus
$$\lim_{\Delta t\to 0}\mathbb{P}(|B_{t+\Delta t}-B_{t}|\ge \delta)=\lim_{\Delta t\to 0}\mathbb{P}(|B_{t+\Delta t}-\mathbb{E}[B_{t+\Delta t}|\mathcal{F}_t]\,|\ge \delta)$$
By application of Chebyshev's inequality, we have
$$\mathbb{P}(|B_{t+\Delta t}-B_{t}|\ge \delta)\le\frac{\operatorname{Var}(B_{t+\Delta t}\,|\mathcal{F}_t)}{\delta^2}\\
\quad\qquad\qquad\qquad\qquad\qquad\qquad=\frac{\operatorname{Var}(B_{t+\Delta t}+B_t-B_t\,|\mathcal{F}_t)}{\delta^2}\\
\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,=\frac{\operatorname{Var}(B_{t+\Delta t}-B_t\,|\mathcal{F}_t)}{\delta^2}+\frac{\operatorname{Var}(B_t\,|\mathcal{F}_t)}{\delta^2}\\
\quad\qquad\quad=\frac{\Delta t}{\delta ^2}$$
since Brownian motion has independent increments .On the other hand $B_t$ is $\mathcal{F}_t$ measurable , hence $\operatorname{Var}(B_t\,|\mathcal{F}_t)=0$. As a result
$$\lim_{\Delta t\to 0}\mathbb{P}(|B_{t+\Delta t}-B_{t}|\ge \delta)=0$$
