I am trying to show the following statement



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> Let $D\subset \mathbb{R}^2$ be an open and **bounded** subset. $\Pi=(P^x : x \in D )$ a Family of standard Brownian Motions started at $x \in D$. Then $\Pi$ ist a tight (or relativly compact due to Prorhorvs Theorem).
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I need that result to proof the existence of a certain type of Process. Unfortunately I do not have any experience working with the concept of "tightness". 



Family of Brownian Motions
Using the fact that $S:=C[0,+\infty):=(f\colon [0,+\infty)\to \Bbb R, f\mbox{ continuous })$ endowed with the metric $$d(f,g):=\sum\_{j=1}^{+\infty}2^{—j}\min(1,\sup\_{0\leq t\leq j}|f(x)-g(x)|)$$ is polish, it's enough to show that each subsequence is tight. Using theorem 4.10 in Karatzas-Shreve book's, we know that a sequence of probability measures $(\mathbb P\_n)\_{n\geq 1}$ over the Borel $\sigma$-albegra is tight if and only if the two following conditions are satisfied: 


1. $\lim\_{\lambda\to+\infty}\sup\_{n\geq 1}\mathbb P\_n(\omega,|\omega(0)|\geq \lambda)=0$ and
2. $\lim\_{\delta\to 0}\sup\_{n\geq 1}\mathbb P\_n(\omega,m^T(\omega,\delta)\geq \varepsilon)=0$ for each $T>0$ and $\varepsilon>0$, where $m^T(\omega,\delta)=\sup\_{|t\_1-t\_2|\leq\delta,0\leq t\_1,t\_2\leq T}|\omega(t\_1)-\omega(t\_2)|$.


We apply this to $\mathbb P\_n=\mathbb P\_{x\_n}$, where $(x\_n)\_{n\geq 1}$ is a sequence in $D$. By boundedness of $D$, the first condition is satisfied. We can control the modulus of continuity using the fact that $$\mathbb P\_n(\omega,m^T(\omega,\delta)\geq \varepsilon)
\leq \frac 1{\varepsilon^2}\delta^2,$$
which is a consequence of the fact that the increments of the Brownian motion $W\_{t\_2}-W\_{t\_1}$ are normally distributed, of normal distribution of mean $0$ and variance $t\_2-t\_1$. 


In fact, with condition 1., we can see that $(\mathcal P\_x,x\in D)$ is tight (where $\mathcal P\_x$ is associated to a Brownian motion started at $x$) if and only if $D$ is bounded. 


