Changing a conditional kernel on a null fiber leaves information unchanged

By [F2], the restriction $P_Z$ is a probability and $P_Z(\{0\})=0$. On each finite target set, define distance 0 on equal points and 1 on distinct points; distinct endpoints force at least one intermediate distance to be 1, so the triangle inequality holds. Tolerance $1\mathrm{/}2$ makes every Cauchy sequence eventually constant; the whole finite set is dense, and radius-$1\mathrm{/}2$ balls are singletons, so the space is complete and discrete, hence Polish and standard Borel by [F1]. The product target is standard Borel by the finite-product supplier in [F1]. For any Borel $C\subseteq [0,1]$ and $A\subseteq \{0,1\}$, replacing ${\delta }_0$ by $K^{\mathrm{\prime }}$ changes ${\int }_{C}K(z,A)P_Z(dz)$ only at $z=0$, a null point. Thus both K and K' satisfy the conditional-law integral identity for X given Z; the constant kernel L satisfies it for Y, and $J(z)={\delta }_{(0,0)}$ satisfies it for $(X,Y)$. These kernels are measurable because their only nonconstant set of values is the Borel singleton $\{0\}$.

Let $J(z)={\delta }_{(0,0)}$ be a conditional law of $(X,Y)$ given Z. With the original kernels, $K(z)\otimes L(z)={\delta }_{(0,0)}$ for every z, so the pointwise KL is zero. With the alternative kernels, the product is ${\delta }_{(1,0)}$ at z=0 and ${\delta }_{(0,0)}$ for z>0. By [F5], $D(J(0)\mathrm{\parallel }{K}^{\mathrm{\prime }}(0)\otimes L^{\mathrm{\prime }}(0))=+\mathrm{\infty }$ because the first law charges $(0,0)$ and the reference gives it mass zero; for z>0 the divergence is zero. Thus the pointwise integrands differ only at a null point, and both conditional-KL integrals in [F4] are zero.

The joint law $P_{ZXY}$ and the conditional-product reference law built from K,L agree. K' equals K $P_Z$-almost everywhere, so [F3] says the reference law built from K',L' is the same law. Consequently both versions give $I(X;Y\mid Z)=D(P_{ZXY}\mathrm{\parallel }{P}_{ZXY})=0$.