Consistency of a family of probabilities in the construction of a pre-brownian motion I'm trying to understand how to construct the pre-brownian motion and I'm stuck at trying to prove the consistency of a family to be able to apply the Kolmogorov extension theorem. 
Definition. Let $S$ be any arbitrary set and $(X,\mathcal{F})$ a events space. Let $\Omega=X^{J}$ and for all $J\subset S$ ($J$ finite) and define $\pi_{J}$ the projection from $\Omega$ to $X_{J}$. Now, suppose that for each $J\subset S$ ($J$ finite) we have a probability measure $P_{J}$ defined over $X^{J}$. We say that $\{P_{J}:J\subset S, J~\text{finite} \}$ is a consistent family if for $J' \subset J$we have that $$P_{J}\Bigl( \pi_{J}\left(\pi_{J'}^{-1}(A) \right) \Bigr)=P_{J'}(A)$$
For $A\in \prod _{i\in J'} \mathcal{F}$.
Now, lets consider the space $\displaystyle\left(\mathbb{R}^{(0,\infty)}, \mathcal{B}(\mathbb{R})^{(0,\infty)} \right)$ and the density function $\displaystyle p(t,x)=\frac1{\sqrt{2\pi t}} e^{\frac{-x^{2}}{2t}}$. 
So for $J=\{t_{1},t_{2},...,t_{n}\}$ and for $A_{1},A_{2},...,A_{n} \in \mathcal{B}(\mathbb{R})$ we define the probability measure
$$P_{J}(A_{1} \times \cdots \times A_{n})=\int_{A_{n}}\int_{A_{n-1}} \cdots \int _{A_{1}} p(t_{1},x_{1})\prod p(t_{j+1}-t_{j}, x_{j+1} - x_{j}) dx_{1}dx_{2} \cdots dx_{n}$$
Where $P_{J}$ is defined over $(\mathbb{R}^{J},\mathcal{B}(\mathbb{R})^{J})$
How can I prove that $P_{J}$ is a consistent family of probability measures? 
Thanks so much for the help.
 A: Say $J = \{t_1,\dots,t_n\}$ and $J' = \{t_{k_1},\dots,t_{k_m}\}$. Fix $A_{k_1},\dots,A_{k_m} \in \mathcal{B}(\mathbb{R})$. Then, $$\pi_J(\pi_{J'}^{-1}(A_{k_1}\times \dots \times A_{k_m})) = \mathbb{R}\times\dots\times\mathbb{R}\times A_{k_1}\times\mathbb{R}\times\dots\times\mathbb{R}\times A_{k_2}\times \mathbb{R}\times\dots.$$ So, with $A := A_{k_1}\times\dots\times A_{k_m}$, we have $$P_J(\pi_J(\pi_{J'}^{-1}(A))) = \int \dots \int p(x_1,t_1)p(x_2-x_1,t_2-t_1)\dots p(x_n-x_{n-1},t_n-t_{n-1})dx_1\dots dx_n.$$ The integrals over $x_1,\dots,x_{k_1-1}$ are over $\mathbb{R}$, as are the integrals over $x_{k_1+1},\dots,x_{k_2-1}$, etc, while the integrals over $x_{k_i}$ are over $A_{k_i}$. So, thinking through the notation for a minute, it suffices to prove the following two general identities: $$ \int_{\mathbb{R}} p(t_1,x_1)p(t_2-t_1,x_2-x_1)dx_1 = p(t_2,x_2)$$ $$\int_{\mathbb{R}} p(t_2-t_1,x_2-x_1)p(t_2-t_2,x_3-x_2)dx_2 = p(t_3-t_1,x_3-x_1).$$ But these are just standard gaussian integrals which can be easily computed via standard formulas.
