For the canonical Brownian motion process on $(C(\mathbb R_{+}), \mathcal{B}, \mu_{0},(X_{t})_{t\geq 0})$ I want to show that. Define for $x \in \mathbb R$, the measure $\mathbb P_{x}$ as the image of $\mu_{0}$ under $C(\mathbb R_{+})\to C(\mathbb R_{+}), \; \gamma \mapsto \gamma +x$. Note that $X_{t}(\gamma)= \gamma(t)$ for $\gamma \in C(\mathbb R_{+})$
I want to show that:
$\mathbb P_{x}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\delta_{x}(A_{0})\int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1})...1_{A_{n}}(x_{n})\;\; (**)$
for $0=t_{0} < ... < t_{n}$ and $A_{0},...,A_{n} \in \mathcal{B}(\mathbb R)$
What has been proven until this point:
$\mu_{0}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\mu_{0}(\{ \gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}\; \forall i = 0,...,n\})\\=\delta_{0}(A_{0}) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1})...1_{A_{n}}(x_{n})\;\ \; \; (*)$
I am struggling to use the $\mu_{0}$ under the translation by $x$.
My attempt:
$\mathbb P_{x}(X_{t_{0}}\in A_{0},...,X_{t_{n}}\in A_{n})=\mu_{0}(\{\gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}-x\; \;\;\forall i = 0,...,n\})$ and by $(*)$ we obtain:
$\mu_{0}(\{\gamma \in C(\mathbb R_{+}): \gamma(t_{i})\in A_{i}-x\; \;\;\forall i = 0,...,n\})=\delta_{0}(A_{0}-x) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}-x}(x_{1})...1_{A_{n}-x}(x_{n})\\=\delta_{x}(A_{0}) \int P_{t_{1}}(x,dx_{1})....P_{t_{n}-t_{n-1}}(x_{n-1},dx_{n})1_{A_{1}}(x_{1}+x)...1_{A_{n}}(x_{n}+x)$
I then thought about substituting $y_{n}=x_{n}+x\implies dy_{n}=dx_{n} $ but realized this does not help either. Any ideas or tips where I am going wrong?
$\textbf{Explanation of my problem:}$ I want to be able to justify why I can conclude $(**)$ from $(*)$. It is clear that I have to work with the image of the measure $\mu_{0}$ under a translation, however, I am struggling as you can see in my above attempt.
The substitution you suggested actually helps. After a slight change of notation, $$ \mu_0(A_0\times A_1\times \cdots\times A_n)=\delta_0(A_0)\int_{A_1} dx_1\cdots\int_{A_n}dx_n\prod_{j=1}^n P_{t_j-t_{j-1}}(x_{j-1},x_j), $$ where $$ P_{t}(x,z)=\frac{1}{\sqrt{2\pi t}}\exp\left(\frac{(x-z)^2}{2t}\right). $$ Using the substitution, $y_j=x_j+x$, and setting $B_j=A_j-x$, \begin{align} \mu_0(B_0\times B_1\times \cdots\times B_n)&=\delta_0(B_0)\int_{B_1} dx_1\cdots\int_{B_n}dx_n\prod_{j=1}^n P_{t_j-t_{j-1}}(x_{j-1},x_j) \\ &=\delta_x(A_0)\int_{A_1} dy_1\cdots\int_{A_n}dy_n\prod_{j=1}^n P_{t_j-t_{j-1}}(y_{j-1},y_j). \end{align}