Mutual-information chain rule without subtracting infinities

Let $K(z,\cdot )$ be a regular conditional law of X given Z and let $Q=P_{ZX}\otimes P_Y$ on $(Z\times X)\times Y$. In the coordinate order $Z,X,Y$, Q has Z marginal $P_Z$ and conditional law $K(z,\cdot )\otimes P_Y$ for $(X,Y)$. On every rectangle $C\times A\times B$, the law formed from this base and kernel has mass ${\int }_{C}K(z,A)P_Y(B)P_Z(dz)=P_{ZX}(C\times A)P_Y(B),$ so it equals Q by rectangle uniqueness.

Apply [F3] to the joint law of $(Z,X,Y)$ and Q using their common Z marginal. With $P_{XY\mathrm{\mid }z}$ denoting the conditional law of $(X,Y)$ given Z=z, this gives $I((Z,X);Y)=D(P_{ZXY}\mathrm{\parallel }{P}_{ZX}\otimes P_Y)=\int D(P_{XY\mathrm{\mid }z}\mathrm{\parallel }{P}_{X\mathrm{\mid }z}\otimes P_Y)P_Z(dz)\mathrm{.}$ The formula is an extended sum of nonnegative KL values.

For $P_Z$-almost every z, disintegrate the two measures in step 2.1 over Y. Their Y marginals are $P_{Y\mathrm{\mid }z}$ and $P_Y$, while their conditional X laws are $P_{X\mathrm{\mid }Y,z}$ and $P_{X\mathrm{\mid }z}$, respectively. [F3] gives $D(P_{XY\mathrm{\mid }z}\mathrm{\parallel }{P}_{X\mathrm{\mid }z}\otimes P_Y)=D(P_{Y\mathrm{\mid }z}\mathrm{\parallel }{P}_Y)+\int D(P_{X\mathrm{\mid }Y=y,z}\mathrm{\parallel }{P}_{X\mathrm{\mid }z})P_{Y\mathrm{\mid }z}(dy)\mathrm{.}$ Applying [F3] with the common Y marginal $P_{Y\mathrm{\mid }z}$ identifies the last integral as $D(P_{XY\mathrm{\mid }z}\mathrm{\parallel }{P}_{X\mathrm{\mid }z}\otimes P_{Y\mathrm{\mid }z})$.

Integrate step 3.1 over $P_Z$. The first term becomes $D(P_{YZ}\mathrm{\parallel }{P}_Y\otimes P_Z)$, which equals $I(Z;Y)$ by the measurable coordinate swap [F6]. Applying [F3] with the common Z marginal of the joint law and its conditional-product reference shows that the second term integrates to $I(X;Y\mid Z)$ as defined in [F2]. All terms are nonnegative KL divergences, so their extended sum is well-defined even when one or more are infinite.