Regular conditional laws are not unique on null conditioning values

Take $\mathrm{\Omega }=(0,1)$ with Borel sigma-algebra and Lebesgue probability, Y(omega)=omega and X(omega)=0. The mass is one by [F3]–[F6], integrating the constant derivative of x on [0,1] and ignoring its null endpoints. The singleton $N=\{1\mathrm{/}2\}$ is Borel and has measure zero: for every positive integer n it is contained in an interval of length $2\mathrm{/}n$, so its measure is at most $2\mathrm{/}n$ and hence zero. Each Dirac section is a probability because for disjoint sets at most one contains its point. For Borel A, $L(y,A)={\mathbf{1}}_A(0)+{\mathbf{1}}_{\{1\mathrm{/}2\}}(y)({\mathbf{1}}_A(1)-{\mathbf{1}}_A(0)),$ a measurable function; K has constant measurable evaluations. Thus both satisfy [F2].

For every $H\in \sigma (Y)$ and Borel A, ${\int }_{H}K(Y,A)dP={\mathbf{1}}_A(0)P(H)=P(H\cap \{X\in A\})$. The difference $L(Y,A)-K(Y,A)$ is bounded in absolute value by ${\mathbf{1}}_N(Y)$, whose integral over H is zero. Therefore L satisfies the same identity, proving [F1] for both kernels. They nevertheless disagree at y=1/2: for the event A={1}, $K(1\mathrm{/}2,A)=0$ whereas $L(1\mathrm{/}2,A)=1$. This is a difference of probability measures at an actual conditioning value in (0,1), not merely outside the range of Y. Their equality outside N is consistent with almost-everywhere uniqueness.