Conditional density formula

By [F2], m is measurable and ${\int }_{T}md\nu =\int pd(\mu \times \nu )=\lambda (E\times T)=1$. For each measurable B, apply [F2] to $p{\mathbf{1}}_{E\times B}$ to obtain $\beta (B)={\int }_{B}md\nu$. Put $Z=\{m=0\}$ and $I=\{m=\mathrm{\infty }\}$. Then $\beta (Z)=0$. Also $n\nu (I)\le \int md\nu =1$ for all $n\ge 1$, so $\nu (I)=0$; the nonnegative integral over this measurable null set is zero, even though m is infinite there, giving $\beta (I)=0$. Therefore $D=T\setminus (Z\cup I)$ is measurable and has full marginal mass. No finite value of m at each point follows from total integrability.

For $A\in \mathcal{S}$, let $a_A(y)={\int }_{A}p(x,y)\mu (dx)$. Tonelli applied to $p{\mathbf{1}}_{A\times T}$ makes $a_A$ measurable. It satisfies $0\le a_A\le m$ everywhere. On D both are finite and the denominator is positive, so $a_A\mathrm{/}m$ is a measurable real function there; reciprocal and multiplication are continuous on this finite positive domain. Pasting the constant $\rho (A)$ off D proves measurable evaluations of K. For each $y\in D$, [F2] ensures that $p(\cdot ,y)$ is measurable and [F3] makes $A\mapsto a_A(y)$ a measure of mass m(y); dividing by this finite positive number gives a probability. Off D, K is the supplied probability $\rho$. This proves [F1], including the empty-event and whole-target evaluations.

For measurable A,B, the contribution of $B\setminus D$ to the K-integral under $\beta$ is zero because $0\le K\le 1$ and $\beta (D^c)=0$. On D the density conversion [F7] and $\beta =md\nu$ give ${\int }_{B}K(y,A)\beta (dy)={\int }_{B\cap D}\frac{a_A(y)}{m(y)}m(y)\nu (dy)={\int }_{B\cap D}{a}_A(y)\nu (dy)\mathrm{.}$ On Z, $a_A=0$ since $a_A\le m$; on I its integral is zero because I is $\nu$-null. Thus the last integral is ${\int }_B{a}_{A}d\nu$, which equals ${\int }_{A\times B}pd(\mu \times \nu )=\lambda (A\times B)$ by [F2]. All cancellation was restricted to finite positive m.

If X,Y have joint law $\lambda$, [F5] applied to the bounded measurable function ${\mathbf{1}}_B(y)K(y,A)$ converts step 3.1 into ${\int }_{\{Y\in B\}}K(Y,A)dP=P(X\in A,Y\in B)\mathrm{.}$ The collection of inverse images $Y^{-1}(B)$ is already a sigma-algebra and is exactly $\sigma (Y)$. These are therefore all required conditioning events. Measurability and probability sections follow from step 2.1 under composition with Y, so [F4] proves the RCD assertion. The proof selected no points, versions or exhaustion: the sigma-finite measures, product density and filler probability are supplied. [F6] is required only when using the library conditional-class existence convention.