Why is $P^{\mu}[B_0 \in \Gamma]=\mu(\Gamma)$ where $P^{\mu}(F)=\int_{\mathbb{R}^d}P^{x}(F)\mu(dx) $ and $P^x$ is defined as $$ P^x(F)=P^0(F-x) $$ and $F-x=\{\omega \in C[0,\infty)^d:\omega(.)+x \in F\}, F \in \mathcal{B}(C[0,\infty)^d),x \in \mathbb{R}^d$.
$P^0$ is the $d$-dimensional Wiener measure on $(C[0,\infty)^d, \mathcal{B}(C[0,\infty)^d))$, and $\mu$ is a probability measure on ($\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d)$)
My attempt : By definition of $P^\mu$ we have $P^\mu(B_0 \in \Gamma)= \int_{\mathbb{R}^d}P^{x}(B_0 \in \Gamma)\mu(dx) $
Since by definition under $P^0$, the coordinate mapping process $B_t(\omega)=\omega(t)$ is with the filtration $\mathcal{F}_t^B$ a $d$-dimensional Brownian motion starting at origin and then using the definition of $P^x$ we can rewrite the above as
$\int_{\mathbb{R}^d}P^{x}(B_0 \in \Gamma)\mu(dx)= \int_{\mathbb{R}^d}P^{0}(\{\{B_0 \in \Gamma)\}-x\} \mu(dx)= \int_{\mathbb{R}^d}P^{0}(\omega \in C[0,\infty)^d:\omega(0)+x \in \Gamma)\mu(dx) $ And now since under $P^0$ ,the coordinate mapping process is a $d$-dimensional Bronwian motion starting at zero we have that $P^0(\omega \in C[0,\infty)^d:\omega(0)=0)=1$ and hence we have
$\int_{\mathbb{R}^d}P^{0}(\omega \in C[0,\infty)^d:\omega(0)+x \in \Gamma)\mu(dx)= \int_{\mathbb{R}^d}\mathrm{1}_x(\Gamma)\mu(dx)=\mu(\Gamma)$
I am not really convinced what I wrote is correct and I would really appreciate if you could help point out the fault in my reasoning and help me solve this little problem.