Finite dimensional distrubutions of Brownian motion I am having some trouble with interpreting the following standard expression for the finite distributions for the brownian motion
$P(B_{t_{1}}\in A_{1}, \ldots B_{t_{n}}\in A_{n})=\int_{A_{1}} f({t_{1}},x_{1}-{x_{0}})dx_{1}\cdots \int_{A_{n}}f({t_{n}-t_{n-1}},x_{n}-{x_{n-1}})dx_{n}$
where $f$ is the centered gaussain density. Given that we know the defining properties of the brownian motion, how do we end up here?
 A: It is enough to prove the following lemma:
Lemma: Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space and $X_1,X_2,X_3: \Omega \to \mathbb{R}$ be independent continuous random variables on this space with density functions $f_1, f_2, f_3: \mathbb{R}\to \mathbb{R}$, respectively. If $A_1, A_2, A_3 \in \mathcal{B}(\mathbb{R})$, the Borel $\sigma$-algebra of $\mathbb{R}$, then
$$\mathbb{P}\big[X_1 \in A_1,X_1+X_2 \in A_2,X_1+X_2+X_3 \in A_3 \big]=\int_{A_1}\int_{A_2}\int_{A_3} f_1 (x) f_2(y-x) f_3(z-y)dzdydx$$
Proof: For $i=1,2,3$, let $\mu_i$ be a measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ given by $\mu_i (A)=\mathbb{P}[X_i \in A]$ and let $\mu=\mu_1 \otimes\mu_2 \otimes \mu_3$ be the product measure on $(\mathbb{R}^3,\mathcal{B}(\mathbb{R}^3))$ . Then,
\begin{align}\mathbb{P}\big[X_1 \in A_1,X_1+X_2 \in A_2,X_1+X_2+X_3 \in A_3 \big] &= \int_{\Omega} I_{\{X_1(\omega)\in A_1\}}I_{\{X_1(\omega)+X_2(\omega)\in A_2\}}I_{\{X_1(\omega)+X_2(\omega)+X_3(\omega)\in A_3\}}d \mathbb{P}(\omega) \\ &= \int_{\mathbb{R}^3}I_{x_1 \in A_1}I_{x_1+x_2 \in A_2}I_{x_1+x_2+x_3 \in A_3} d\mu \\ &= \int_{\mathbb{R}^3}I_{x_1 \in A_1}I_{x_1+x_2 \in A_2}I_{x_1+x_2+x_3 \in A_3}f_1(x_1)f_2(x_2)f_3(x_3) dx_1dx_2dx_3\end{align}
Where the second equality is due to the Change of Variables Theorem and the last one is due to the independence of the random variables. If we set $x=x_1$, $y=x_1+x_2$, $z=x_1+x_2+x_3$ and change variables again (note that the Jacobian determinant is $1$), we get the required expression.
Obs: I only did it for $n=3$ but you can easily generalize this argument.
