Simultaneous rational conditional distribution function versions

We first repair the integral interface. Given two finite disjoint simple displays $s={\sum }_i{c}_i{\mathbf{1}}_{E_i}={\sum }_j{d}_j{\mathbf{1}}_{F_j}$, adjoin $E_0=\mathrm{\Omega }\setminus {\bigcup }_i{E}_i$ and $F_0=\mathrm{\Omega }\setminus {\bigcup }_j{F}_j$, both with coefficient zero. The intersections $E_i\cap F_j$, now including indices zero, form a finite measurable partition of all of $\mathrm{\Omega }$. On every nonempty cell the two coefficients agree, so finite additivity gives equal coefficient sums. This includes infinite-measure zero cells because both corresponding products are the stipulated $0\cdot (+\mathrm{\infty })=0$. Thus the simple integral in [F1] is representation-independent. A common zero-complement refinement now proves monotonicity and additivity for simple functions. For a scalar $a>0$, termwise multiplication proves positive homogeneity; for $a=0$, the zero display gives $\int 0=0$ directly, without forming $0\cdot (+\mathrm{\infty })$.

The supremum definition in [F1] therefore gives monotonicity of the nonnegative integral and agreement with the simple integral. Multiplication by $a>0$ bijects the simple minorants of $f$ and $af$, giving positive homogeneity; again $\int 0=0$ is separate. If $0\le f_n\uparrow f$, let $L={\sup }_n\int f_n$. For a simple $s\le f$ and $0<a<1$, put $A_n=\{f_n\ge as\}$. Then $A_n\uparrow \mathrm{\Omega }$. The formula $A\mapsto {\int }_{A}s$ is a measure by the finite partition formula from step 1.1, so [F4] gives ${\int }_{A_n}s\uparrow \int s$. Since $as{\mathbf{1}}_{A_n}\le f_n$, positive homogeneity gives $a{\int }_{A_n}s\le L$; let first $n$ and then $a\uparrow 1$ to obtain $\int s\le L$. Taking the supremum over $s$ proves monotone convergence. Using the approximants in [F2], simple additivity and monotone convergence give $\int (f+g)=\int f+\int g$. Consequently $A\mapsto {\int }_{A}f$ is a measure: apply monotone convergence to finite unions of a disjoint sequence. If the ambient measure is finite and $0\le f_n\downarrow f\le M$, then $M-f_n\uparrow M-f$; additivity in the finite identity $f_n+(M-f_n)=M$ proves bounded decreasing convergence.

We next construct conditional densities without importing a conditional-expectation theorem. For an event $C\in \mathcal{F}$, set ${\nu }_C(A)=P(A\cap C)$ on $\mathcal{G}$. Let $\mathcal{H}$ be the nonnegative measurable $h$ satisfying ${\int }_{A}hdP\le {\nu }_C(A)$ for every $A\in \mathcal{G}$, and let $M={\sup }_{h\in \mathcal{H}}\int hdP\le 1$. The maximum of two members remains in $\mathcal{H}$: split each test event over $\{h\ge k\}$ and its complement and use step 2.1. By [F5] choose $h_j\in \mathcal{H}$ with integrals tending to $M$, and put $f_j={\max }_{i\le j}{h}_i\uparrow f$. By F8 the supremum f is measurable. Step 2.1 gives $f\in \mathcal{H}$ and $\int f=M$. The set function $\rho (A)={\nu }_C(A)-{\int }_{A}fdP$ is a finite positive measure. If it were not singular to $P{\mathrm{\mid }}_{\mathcal{G}}$, choose by [F3] a positive set $Q_m$ for $\rho -P\mathrm{/}m$ for every $m\ge 1$. If every $P(Q_m)=0$, their union is null by [F7], and on its complement $\rho (A)\le P(A)\mathrm{/}m$ for every $m$, so $\rho$ is concentrated on a $P$-null set, a contradiction. Hence some $Q_m$ has positive $P$-measure, and $f+m^{-1}{\mathbf{1}}_{Q_m}\in \mathcal{H}$ has integral greater than $M$, again a contradiction. Thus $\rho \perp P$. Since $0\le \rho \le {\nu }_C\le P$, it is also absolutely continuous, so its singular carrier immediately gives $\rho =0$. We have proved ${\nu }_C(A)={\int }_{A}fdP$ for all $A\in \mathcal{G}$.

Since $\int f\le 1$, the sets $\{f\ge n\}$ show that $f<\mathrm{\infty }$ almost surely. Comparing ${\nu }_C\le P$ on $\{f\ge 1+1\mathrm{/}m\}$ shows that $f\le 1$ almost surely. Changing $f$ on the measurable union of these null sets gives a real $[0,1]$-valued density. If $f,g$ represent ${\nu }_C,{\nu }_D$ and $C\subseteq D$, then on $A_m=\{f\ge g+1\mathrm{/}m\}$, step 2.1 gives ${\int }_{A_m}f\ge {\int }_{A_m}g+P(A_m)\mathrm{/}m,$ whereas the representation identities give ${\int }_{A_m}f\le {\int }_{A_m}g$. Hence $P(A_m)=0$ for every $m$, so $f\le g$ almost surely. Taking $C=D$ in both directions proves uniqueness; $C=\mathrm{\varnothing },\mathrm{\Omega }$ proves the constant-zero and constant-one assertions. For $C_j\uparrow C$, set all selected densities to zero on the countable union of their order failures and put $u={\lim }_j{f}_j$. Step 2.1 and [F4] give ${\int }_{A}u={\lim }_{j}P(A\cap C_j)=P(A\cap C)$, so uniqueness identifies $u$ with the density of $C$. Decreasing event limits follow by applying this to complements and using ${\int }_A(1-f)=P(A)-{\int }_{A}f$, an instance of the finite additivity from step 2.1. This proves every restricted conditional-density assertion in the statement.

Apply step 3.1 to $C_q=\{X\le q\}$ and use [F5] to select one real $\mathcal{G}$-measurable density $G_q$ for each rational $q$. The selection is countable by [F6]. Step 4.1 gives $0\le G_q\le 1$ almost surely and $G_q\le G_r$ almost surely whenever $q<r$. Since $C_n\uparrow \mathrm{\Omega }$, $C_{-n}\downarrow \mathrm{\varnothing }$, and $C_{q+1\mathrm{/}n}\downarrow C_q$, the event-limit clause of step 4.1 gives the two tail limits and every rational right-limit almost surely.

Form the union $N$ of the failure sets for the bounds, rational order comparisons, two tail limits, and the right-limit assertion for each rational $q$. They belong to $\mathcal{G}$ by rational-cut descriptions and [F8], and they are null by step 5.1. There are countably many by [F6], so [F7] gives $P(N)=0$. Replace each $G_q$ on $N$ by ${\mathbf{1}}_{\{q\ge 0\}}$. Pasting preserves measurability. A bounded difference supported on N has absolute value at most a constant times ${\mathbf{1}}_N$, whose integral is zero by step 2.1, so all event integrals are unchanged and the functions remain versions. The filler is increasing, has the required tails, and satisfies the right-limit property also at $q=0$.