Regular conditional law for a finite partition

For a positive-mass cell, preimages under X preserve disjoint unions, so $A\mapsto P(H_j\cap X^{-1}(A))$ is countably additive, vanishes at the empty set and has total mass $P(H_j)$. Division by this positive finite mass gives a probability. On zero-mass cells the supplied $\rho$ is a probability. For each A the evaluation is constant on every cell and is therefore measurable for the finite partition sigma-algebra. Every event H in that sigma-algebra is a union of cells: the set of such unions is itself a sigma-algebra containing the cells. Thus ${\int }_{H}K(\omega ,A)dP={\sum }_{j:H_j\subseteq H}P(H_j)K{\mathrm{\mid }}_{H_j}(A)={\sum }_{j:H_j\subseteq H}P(H_j\cap \{X\in A\})=P(H\cap \{X\in A\})\mathrm{.}$ Each zero cell contributes zero on both sides, and an empty cell can be ignored. This proves [F1]–[F2].

In the displayed finite model the cell masses are $1\mathrm{/}2,1\mathrm{/}2,0$. On the first cell, $P(X=0,H_1)=1\mathrm{/}4$ and $P(X=1,H_1)=1\mathrm{/}4$, so the conditional probabilities are $1\mathrm{/}2$ and $1\mathrm{/}2$. On the second cell, $P(X=1,H_2)=1\mathrm{/}2$ gives probability one at 1. The third uses the specified filler despite $X(4)=2$, since its entire cell has mass zero. For instance testing A={1} and H=Omega gives $(1\mathrm{/}2)(1\mathrm{/}2)+(1\mathrm{/}2)(1)+0(0)=3\mathrm{/}4=P(X=1)$; testing H=$H_1$ gives $1\mathrm{/}4$ on both sides. Hence the calculated kernels have exactly the claimed values.