The transition probability density function of the standard Wiener process is: $$ f(x_2,t_2|x_1,t_1) = \frac{1}{\sqrt{2 \pi (t_2-t_1)}}e^{-\frac{(x_2-x_1)^2}{2(t_2-t_1)^2}} $$
I know that if Markov process is continuous, then its transition probability density function satisfies, $\forall \epsilon > 0$ and uniformly in $x_1$, $t$, and $\Delta t$ $$ \text{lim}_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{|x_2-x_1|>\epsilon} f(x_2, t+\Delta t|x_1, t) dx_2 = 0 $$
I would like to prove it for the standard Wiener process above. If anyone has an idea that would be very appreciated
First note that $$f(x_2,t+\Delta t \ | \ x_1,t) = \frac{1}{\sqrt{2\pi \Delta t}} \exp\left\{\frac{-1}{2}\frac{(x_2-x_1)^2}{\Delta t}\right\},$$ Hence we can write (let $u = x_2-x_1$) \begin{align} &\int_{|x_2-x_1|>\epsilon} \frac{1}{\sqrt{2\pi \Delta t}} \exp\left\{\frac{-1}{2}\frac{(x_2-x_1)^2}{\Delta t}\right\}dx_2 \\ &= \int_{|u|>\epsilon} \frac{1}{\sqrt{2 \pi \Delta t}} \exp\left\{\frac{-u^2}{2\Delta t}\right\}du \\ &= 2 \int_{\epsilon}^{\infty} \frac{1}{\sqrt{2 \pi \Delta t}} \exp\left\{\frac{-u^2}{2\Delta t}\right\}du. \\ \end{align} Let $u/\sqrt{\Delta t} = w \Rightarrow du = dw \sqrt{\Delta }$ and hence the above integral reduces to \begin{align} &2 \int_{\frac{\epsilon}{\sqrt{\Delta t}}}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left\{\frac{-w^2}{2}\right\}dw. \\ &= 2\left(1-\Phi\left(\frac{\epsilon}{\sqrt{\Delta t}}\right)\right), \end{align} where $\Phi$ is the standard normal cdf. Hence, \begin{align} &\lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x_2-x_1|>\epsilon} f(x_2,t+\Delta t \ | \ x_1,t)dx_2 \\ &= \lim_{\Delta t \to 0} \frac{2}{\Delta t} \left(1-\Phi\left(\frac{\epsilon}{\sqrt{\Delta t}}\right)\right) \quad (*) \end{align} You can prove that $(*)$ is indeed equal to zero. One possibility is to use l'Hopital's rule. Another possibility is an application of Markov's inequality.