Let $(\Omega,\mathcal{A},\mathbb{P})$ a probability space and $(B_t)_{t \geq 0}$ a Brownian motion (started in $x=0$). Then one can define a probability measure $\mathbb{P}^x$, $x \in \mathbb{R}$, on $(\Omega,\sigma(B_t; t \geq 0))$ by
$$\mathbb{P}^x(B_{t_1} \in A_1, \ldots, B_{t_n} \in A_n) := \mathbb{P}(x+B_{t_1} \in A_1, \ldots, x+B_{t_n} \in A_n)$$ where $A_j \in \mathcal{B}(\mathbb{R})$, $t_j \geq 0$.
This implies in particular $\mathbb{P}^x(B_0=x)=\mathbb{P}(B_0=0)=1$, i.e. $B_0 \sim \delta_x$ with respect to $\mathbb{P}^x$.
But what's wrong about the following argumentation? Assume that $B_0(w)=0$ for all $w \in \Omega$. Then $B_0(\cdot)^{-1}(\{x\})=\emptyset$ for all $x \not= 0$, hence in particular $\mu(B_0=x)=0$ for any measure $\mu$. This clearly contradicts $\mathbb{P}^x(B_0=x)=1$.
Thanks!
This remark is a good reason to avoid assuming that $B_0(\omega)=0$ for all $\omega$ in $\Omega$. Instead, one usually defines different measures $\mathbb P^x$ on a common probability space $\Omega$, and each $\mathbb P^x$ sees a different part of $\Omega$.
For example, one can choose for $\Omega$ the space of continuous functions $[0,+\infty)\to\mathbb R$ and define $(B_t)_{t\geqslant0}$ by $B_t:\omega\mapsto\omega(t)$, for every $t\geqslant0$. Then each $\mathbb P^x$ is a probability measure on the entire $\Omega$ but $\mathbb P^x(\Omega_x)=1$ for every $x$, where $\Omega_x=[B_0=x]$, that is, $\Omega_x\subset\Omega$ is the space of all the continuous functions $\omega:[0,+\infty)\to\mathbb R$ such that $\omega(0)=x$ (note that each $\Omega_x$ is measurable). In particular, $\mathbb P^x$ and $\mathbb P^y$ are mutually singular for every $x\ne y$ since $\Omega_x\cap\Omega_y=\varnothing$.