Brownian finite-dimensional density

For each $j$, let ${\sigma }_j=\sqrt{h_j}>0$ and $g_j(y)=\frac{1}{{\sigma }_j}\varphi (y\mathrm{/}{\sigma }_j)=\frac{1}{\sqrt{2\pi h_j}}{e}^{-y^2\mathrm{/}(2h_j)}\mathrm{.}$ For a Borel set $A\subseteq \mathbb{R}$, apply [F6] to the dilation $T_j(z)={\sigma }_{j}z$ and the nonnegative Borel function ${\mathbf{1}}_A{g}_j$. Since $T_j^{\mathrm{\prime }}={\sigma }_j$, this gives ${\int }_A{g}_j(y)dy={\int }_{\mathbb{R}}{\mathbf{1}}_A({\sigma }_{j}z)g_j({\sigma }_{j}z){\sigma }_{j}dz={\int }_{\mathbb{R}}{\mathbf{1}}_A({\sigma }_{j}z)\varphi (z)dz\mathrm{.}$ By the pushforward definition in [F2], $g_j$ is therefore a density of $N(0,h_j)$.

Define $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ and $S:{\mathbb{R}}^n\to {\mathbb{R}}^n$ by $(Ty{)}_j={\sum }_{k=1}^j{y}_k,(Sx{)}_j=x_j-x_{j-1},x_0=0.$ Direct telescoping gives $ST=TS=\mathrm{id}$. Their coordinate partial derivatives are constant, so [F7] makes them $C^1$ with their displayed matrices. The matrix of $S$ is lower triangular with every diagonal entry one; hence $\mathrm{\mid }\det DS\mathrm{\mid }=1$. Thus $S$ is a $C^1$ diffeomorphism with inverse $T$. Moreover, $T\mathrm{\Delta }=(B_{t_1},\dots ,B_{t_n})$ almost surely by telescoping and $B_0=0$ almost surely.

Put $q(y_1,\dots ,y_n)={\prod }_{j=1}^n{g}_j(y_j)$. Induction on $n$, using Tonelli and the Borel equality ${\lambda }_{n-1}\times {\lambda }_1={\lambda }_n$, shows that the measure $Q(E)={\int }_{E}qd{\lambda }_n$ has on every Borel rectangle the value ${\prod }_j{\int }_{A_j}{g}_{j}d{\lambda }_1$. The base $n=1$ is step 1.1; the induction also gives $Q({\mathbb{R}}^n)=1$. Thus [F4] and uniqueness of the product measure identify $Q$ with the product of the $N(0,h_j)$ laws. By [F1] and [F3], $Q$ is exactly the joint law of $\mathrm{\Delta }=({\mathrm{\Delta }}_1,\dots ,{\mathrm{\Delta }}_n)$.

For a Borel set $A\subseteq {\mathbb{R}}^n$, step 2.1 and the almost-sure identity in step 1.2 give $P((B_{t_1},\dots ,B_{t_n})\in A)={\int }_{T^{-1}A}q(y)dy\mathrm{.}$ Apply [F6] to $S$ and $f(y)={\mathbf{1}}_{T^{-1}A}(y)q(y)$. Because $T(Sx)=x$ and $\mathrm{\mid }\det DS\mathrm{\mid }=1$, the right side becomes ${\int }_{{\mathbb{R}}^n}{\mathbf{1}}_A(x)q(Sx)dx={\int }_A{\prod }_{j=1}^n{g}_j(x_j-x_{j-1})dx\mathrm{.}$ Substituting the formula from step 1.1 is exactly the stated density.

The strict inequalities make every $h_j$ positive, so no division by zero or singular normal density occurs. For $n=1$, the formula is the $N(0,t_1)$ density with $x_0=t_0=0$; the empty case $n=0$ is excluded. The triangular determinant is one even when $n=1$. AC is used through the Brownian and normal-law suppliers, and it supplies the countable choice required by [F5] and [F6]; the finite triangular transformation makes no additional choice.