Regular conditional law of one coordinate given another

The triangle is product-measurable, since its defining strict inequalities between coordinates are open conditions (or countable rational rectangle unions). For $0<y<1$ the section integral is $m(y)={\int }_0^{y}2dx=2y$, and it is zero for all other y. The constant primitive 2x computes this integral by [F2] and [F3]; finite endpoints are Lebesgue-null, as follows from containment in intervals of arbitrarily small length. Likewise ${\int }_0^{1}2ydy=[y^2{]}_0^1=1$. Thus [F5] shows the nonnegative joint density has total mass one, and [F6] constructs the probability. The bridge uses the countable choice supplied by [F4].

On $0<y<1$, $0<m(y)=2y<\mathrm{\infty }$ and ${\int }_{A}p(x,y)dx=2{\lambda }_1(A\cap (0,y))$. Dividing gives the displayed uniform kernel. Off this interval the marginal is zero, so [F1] allows the supplied point-mass probability ${\delta }_0$. Its event value is ${\mathbf{1}}_A(0)$ and it is countably additive because at most one member of a disjoint event sequence contains 0. For any Borel A,B, the conditional rectangle calculation is ${\int }_{B}K(y,A)P_Y(dy)={\int }_{B\cap (0,1)}\frac{{\lambda }_1(A\cap (0,y))}{y}2ydy={\int }_{B\cap (0,1)}2{\lambda }_1(A\cap (0,y))dy,$ the joint probability by [F5]. In particular $K(1\mathrm{/}2,(0,1\mathrm{/}4))=(1\mathrm{/}4)\mathrm{/}(1\mathrm{/}2)=1\mathrm{/}2$, while $K(0,(0,1\mathrm{/}4))=0$ under the specified filler. The latter is not a value of a density ratio.