Given a Probability space $(\Omega,{\mathcal B}(\Omega),P)$, where $\Omega$ is a Polish space and equipped with the distance$$d:\Omega\times \Omega \rightarrow [0,\infty).$$${\mathcal B}(\Omega)$ is the Borel $\sigma$-algebra of $\Omega$. $\Omega_0 \subset \Omega$ with $P(\Omega_0)=1$. Knowing that, equipped with the same distance function $$d_0:\Omega_0\times \Omega_0 \rightarrow [0,\infty) \quad d_0(x,y)=d(x,y)\quad x,y\in\Omega_0.$$$\Omega_0$ is also a separable metric space.
My question is, whether can we get ${\mathcal B}(\Omega)\bigcap \Omega_0={\mathcal B}(\Omega_0)$?
In fact, let\begin{equation}B(x,\delta)=\{y\in \Omega,d(y,x)<\delta\} \quad x\in\Omega,\delta>0,\end{equation}and\begin{equation}B_0(x,\delta)=\{y\in \Omega_0,d_0(y,x)<\delta\} \quad x\in\Omega_0,\delta>0.\end{equation}We have ${\mathcal B}(\Omega)\bigcap \Omega_0=\sigma(\{B(x,\delta)\bigcap\Omega_0,x\in\Omega,\delta>0\})$, respectively ${\mathcal B}(\Omega_0)=\sigma(\{B_0(x,\delta),x\in\Omega_0,\delta>0\})$.
It can be easy seen $B_0(x,\delta)=B(x,\delta)\bigcap \Omega_0$ for $x\in\Omega_0$ so that ${\mathcal B}(\Omega)\bigcap \Omega_0\supset{\mathcal B}(\Omega_0)$, but could we get ${\mathcal B}(\Omega)\bigcap \Omega_0\subset{\mathcal B}(\Omega_0)$? Or point out the inadequacy in my statement.