Conditional Distributions and Regular Conditional Probability

First suppose K finite and write $m(s)=K(s,T)<\mathrm{\infty }$. For a product-measurable set E, its section $E_s$ is measurable, by applying the section theorem to its indicator. Let $\mathcal{D}$ consist of those E for which $s\mapsto K(s,E_s)$ is measurable. It contains every rectangle $A\times B$, whose evaluation is $1_A(s)K(s,B)$, and it contains $S\times T$. If E belongs to this class, then $(E^c{)}_s=T\setminus E_s$ and $K(s,(E^c{)}_s)=m(s)-K(s,E_s)$; both terms are finite measurable real functions, so the difference is measurable. If E_j are disjoint class members, their sections are disjoint and $K(s,({\bigcup }_j{E}_j{)}_s)={\sum }_{j}K(s,(E_j{)}_s)$ is an increasing limit of measurable finite sums. Thus this class is a lambda-system. Rectangles form a pi-system, so Dynkin's theorem gives every product-measurable E in the class. The measurable-closure proposition can be used on S equipped with its zero measure; its conclusions concern only Sigma and do not require a preexisting source probability.

For a nonnegative simple product-measurable function $g={\sum }_{j=1}^r{a}_j{1}_{E_j}$ with disjoint E_j and finite nonnegative coefficients, its section integral is ${\sum }_j{a}_{j}K(s,(E_j{)}_s)$ and is measurable by step 1.1. Choose the prescribed increasing simple approximation $g_n\uparrow f$ on the product. For every s the section theorem and monotone convergence give $I_f(s)={\lim }_n{I}_{g_n}(s)$, so the increasing-limit closure proves measurability. No uniform bound in s was used; only the individual finite masses entered the complement calculation.

Now let $(T_n)$ be the specified common exhaustion. Define $K_n(s,A)=K(s,A\cap T_n)$. For each s this is the restriction of a measure, its evaluations are measurable by the kernel hypothesis, and $K_n(s,T)=K(s,T_n)<\mathrm{\infty }$. Apply step 2.1 to K_n. Sectionwise integration against the restriction equals integration of $f(s,\cdot )1_{T_n}$ against K(s,·): this holds for indicators by definition, for simple functions by finite sums, and for nonnegative functions by monotone convergence. Since $T_n\uparrow T$, $I_f(s)={\lim }_n\int f(s,t)K_n(s,dt)$; another increasing-limit argument proves the result. If T is empty all these integrals are zero; if S is empty the assertion is vacuous.

For real f its positive and negative parts and absolute value are product-measurable. The proved result makes $I_{f^{+}},I_{f^{-}},I_{\mathrm{\mid }f\mathrm{\mid }}$ measurable. Therefore $D={\bigcup }_{n\ge 1}\{I_{\mathrm{\mid }f\mathrm{\mid }}<n\}$ is measurable. On D both part integrals are finite since each is at most $I_{\mathrm{\mid }f\mathrm{\mid }}$. Define $J_{+}(s)=I_{f^{+}}(s)$ on D and zero otherwise, and define J_- similarly. Nonnegative measurable restriction makes both J_± measurable; they are finite everywhere. Their real difference is the requested signed integral on D and zero elsewhere. This never subtracts two infinities. For a zero kernel all integrals vanish and D=S, even if f is unbounded; for f=0 the same holds for every permitted kernel. All approximations and the exhaustion are specified; no AC is used.

For each measurable $A\subseteq U$, the function $L(\cdot ,A)$ is measurable and between zero and one. Its lift to $S\times T$ is product-measurable because inverse images are rectangles with first factor S. The integration theorem therefore gives measurable source evaluations of KL. For a fixed s, $(KL)(s,\mathrm{\varnothing })=0$ and $(KL)(s,U)=\int 1dK(s,\cdot )=1$. If $(A_j{)}_{j\in \mathbb{N}}$ are disjoint measurable subsets of U, then $L(t,{\bigcup }_{j\in \mathbb{N}}{A}_j)={\lim }_N{\sum }_{j=0}^{N}L(t,A_j)$, an increasing limit. Monotone convergence and finite additivity of the integral yield $(KL)(s,{\bigcup }_j{A}_j)={\sum }_j(KL)(s,A_j)$. Hence each source section is a probability measure.

Fix s. For every nonnegative measurable $h:U\to [0,\mathrm{\infty }]$, ${\int }_{U}h(u)(KL)(s,du)={\int }_T({\int }_{U}h(u)L(t,du))K(s,dt)\mathrm{.}$ For h=1_A this is exactly the definition. Finite nonnegative linear combinations establish it for simple h. For increasing simple h_n converging to h, monotone convergence first under each L(t,·), then under K(s,·), and also under (KL)(s,·), proves the identity. The inner integrals are measurable by the integration theorem, so every displayed integral is defined. This is an iterated-kernel identity proved locally; it is not an application of product-measure Fubini to a varying measure.

For $A\subseteq V$ measurable take the bounded measurable function $h(u)=M(u,A)$ in step 2.1. Its left side is $((KL)M)(s,A)$, while its right side is ${\int }_T(LM)(t,A)K(s,dt)=(K(LM))(s,A)$. Step 1.1 also ensures LM is a probability kernel, so both compositions are defined. This proves the asserted pointwise equality for every s and A. When A is empty both sides are zero; when A=V both are one. For an empty source equality is vacuous. If an intermediate target is empty while its source is nonempty, the hypothesized probability kernel cannot exist; no measure of mass one on the empty set is used. The proof requires no AC.

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 $q=0$.

Use the everywhere good filled data of [F1], writing them again $G_q$. For fixed $\omega$, put $F_{\omega }(x)={\inf }_{q\in \mathbb{Q},q>x}{G}_q(\omega )$. The index set is nonempty by F7 and countable by F8. Its values lie in $[0,1]$, and shrinking it as $x$ increases shows monotonicity. For rational $r$, monotonicity of the data gives $G_r\le F_{\omega }(r)\le G_{r+1\mathrm{/}n}$ for every positive integer $n$, so the rational right-limit condition gives $F_{\omega }(r)=G_r$. In particular $F_{\omega }(n)\to 1$ and $F_{\omega }(-n)\to 0$, and monotonicity gives both real tail limits.

If $x_j\downarrow x$, monotonicity gives a limit $L\ge F_{\omega }(x)$. For every rational $q>x$, eventually $x_j<q$, so $F_{\omega }(x_j)\le G_q(\omega )$ and $L\le G_q(\omega )$. Taking the infimum gives $L\le F_{\omega }(x)$. In particular the sequence $x+1\mathrm{/}j$ proves right continuity. By the existence clause of [F2] there is a Borel probability with CDF $F_{\omega }$. If $\mu ,\nu$ are two such probabilities, the equality class $\mathcal{C}=\{A:\mu (A)=\nu (A)\}$ contains $\mathbb{R}$ and every closed half-line, is closed under complements by subtraction from the common finite total one, and is closed under countable disjoint unions by countable additivity. It is a lambda-system containing the closed-half-line pi-system, whose generated sigma-algebra is Borel by [F4]; [F3] gives $\mu =\nu$. Thus the probability is locally proved unique and defines $K(\omega ,\cdot )$ without a family choice. On $N$ its CDF is ${\mathbf{1}}_{\{x\ge 0\}}$, so the same uniqueness identifies it with ${\delta }_0$.

For fixed real $x$, [F5] makes $F_{\omega }(x)$ a $\mathcal{G}$-measurable countable infimum. Using F7 and the fixed enumeration in F8, choose recursively the first enumerated rational $q_1\in (x,x+1)$ and then the first $q_{j+1}\in (x,\min (q_j,x+1\mathrm{/}(j+1)))$; thus $q_j\downarrow x$. Step 2.1 and rational agreement give $G_{q_j}\downarrow F_{\omega }(x)$. The bounded decreasing-convergence clause locally proved in [F1], applied on $H$, gives ${\int }_H{F}_{\omega }(x)dP={\lim }_j{\int }_H{G}_{q_j}dP={\lim }_{j}P(H\cap \{X\le q_j\})=P(H\cap \{X\le x\})\mathrm{.}$ The last indicator limit includes the point $X=x$.

Let $\mathcal{D}$ be the Borel sets $A$ for which $\omega \mapsto K(\omega ,A)$ is $\mathcal{G}$-measurable and the stated identity holds for every $H\in \mathcal{G}$. It contains $\mathbb{R}$. Complements use $1-K(\omega ,A)$ and subtraction from the finite total $P(H)$. For pairwise disjoint $A_j\in \mathcal{D}$, countable additivity of each section gives the union evaluation as the increasing limit of finite partial sums. This is measurable by [F5], and the locally proved monotone convergence in [F1] gives the event identity. Hence $\mathcal{D}$ is a lambda-system containing the closed-half-line pi-system from step 3.1. By [F3] and [F4] it contains every Borel set. The measurable evaluations and probability sections prove both kernel axioms and the full conditional identity.

The probability space is nonempty, so the existence of $X$ makes $E$ nonempty. By [F3] fix a Borel isomorphism $h:E\to P$, where $P$ has a complete compatible metric $\rho$ and a countable dense set. Use [F9] to enumerate that set as $(p_n)$ and put $d=\min (1,\rho )$. A $d$-Cauchy sequence is eventually at $d$-distance below one, hence is $\rho$-Cauchy; its $\rho$-limit is also its $d$-limit. Thus $d$ is complete and compatible. Define $e(x)=(d(x,p_n){)}_{n\ge 0}\in Q=[0,1{]}^{\mathbb{N}}\mathrm{.}$ Each coordinate is one-Lipschitz. If $x\mathrm{\ne }y$, choose $p_n$ with $d(x,p_n)<d(x,y)\mathrm{/}3$; the reverse triangle inequality makes the $n$th distances different, so $e$ is injective. It is continuous by the initial description of the product topology. Its inverse on $e[P]$ is continuous: for $\epsilon <1$, choose $p_n$ with $d(x,p_n)<\epsilon \mathrm{/}4$; if $z=e(y)$ and $\mathrm{\mid }{z}_n-e(x{)}_n\mathrm{\mid }<\epsilon \mathrm{/}2$, then $d(x,y)<\epsilon$. Hence $e$ is a homeomorphism onto its image, with the inverse-continuity estimate explicit.

On $Q$ set $D(u,v)={\sum }_{n\ge 0}{2}^{-(n+1)}\mathrm{\mid }{u}_n-v_n\mathrm{\mid }$. The geometric tail bound makes this finite; termwise separation and the triangle inequality make it a metric. A $D$-ball controls every prescribed finite set of coordinates because $\mathrm{\mid }{u}_n-v_n\mathrm{\mid }\le 2^{n+1}D(u,v)$. Conversely, after choosing $N$ with ${\sum }_{n\ge N}{2}^{-(n+1)}<\epsilon \mathrm{/}2$, sufficiently small restrictions on the first $N$ coordinates force $D<\epsilon$. Thus $D$ induces the product topology. A $D$-Cauchy sequence is Cauchy in every coordinate, whose limit lies in $[0,1]$ by completeness of the reals; a finite-head plus geometric-tail estimate proves convergence in $D$. Therefore $D$ is complete. The image $Y=e[P]$ is completely metrizable by transport of $d$. AC supplies DC by choosing a successor for every admissible finite history and iterating, so [F5] makes $Y$ a $G_{\delta }$, hence Borel, subset of $(Q,D)$.

Let $a:Q\to C\subseteq [0,1]$ be [F6]. Since $a^{-1}$ is measurable and $Y$ is Borel, $B=a[Y]=(a^{-1}{)}^{-1}[Y]$ is Borel in $C$ and hence in $[0,1]$. Restricting $a$ and its inverse shows that $c=a\circ e\circ h:E\to B$ is bimeasurable. For $q\in \mathbb{Q}$, set $H_q=c^{-1}[B\cap (-\mathrm{\infty },q]]$. Let ${\mathcal{A}}_n$ be the finite Boolean algebra generated by the first $n$ rational cuts in a fixed enumeration and $\mathcal{A}={\bigcup }_n{\mathcal{A}}_n$. By [F8] and [F9], $\mathcal{A}$ is countable; it is an algebra and generates $\mathcal{S}$ by [F7] and bimeasurability. Rational separation and injectivity of $c$ show that it separates points. If finite measures $\mu ,\nu$ agree on $\mathcal{A}$, their equality class is a lambda-system: complements subtract from their common finite total and disjoint unions use countable additivity. It contains the pi-system $\mathcal{A}$, so [F7] gives equality on $\mathcal{S}$. This proves the two auxiliary conclusions without using either affected published standard-Borel interface.

Fix $e_0\in E$ and put $Z=c\circ X$. By [F2] it has an everywhere real conditional probability kernel $\nu$. The evaluation $b(\omega )=\nu (\omega ,B)$ is $\mathcal{G}$-measurable and $0\le b\le 1$. Since $Z\in B$, its conditional identity gives $\int bdP=1$. The repaired finite-additivity interface in [F2] gives $\int (1-b)=0$; for each integer $k\ge 1$, on $N_k=\{1-b\ge 1\mathrm{/}k\}$ monotonicity gives $P(N_k)\mathrm{/}k\le 0$. Thus $N=\{b\mathrm{\ne }1\}={\bigcup }_{k\ge 1}{N}_k$ is measurable and null. Define $K(\omega ,A)=\{\begin{array}{ll}\nu (\omega ,c[A]),& \omega \mathrm{\notin }N,\\ {\mathbf{1}}_A(e_0),& \omega \in N\mathrm{.}\end{array}$ Bimeasurability makes $c[A]$ Borel in the real line, so every evaluation is measurable. Off $N$, injectivity and $\nu (\omega ,B)=1$ give a probability on $E$; on $N$ the filler is Dirac. The two evaluations differ only on a measurable null set, and the local null-integral clause in [F2] gives ${\int }_{H}K(\omega ,A)dP=P(H\cap \{c(X)\in c[A]\})=P(H\cap \{X\in A\})\mathrm{.}$ By [F1], $K$ is the required everywhere regular conditional distribution.

Fix the countable determining algebra $\mathcal{A}$ from [F2], whose AC hypothesis is [F4]. For $A\in \mathcal{A}$, put $u=K(\cdot ,A)$ and $v=L(\cdot ,A)$. By [F1], both are $[0,1]$-valued and have equal integrals on every $H\in \mathcal{G}$. For $m\ge 1$, the set $H_m^{+}=\{u\ge v+1\mathrm{/}m\}$ belongs to $\mathcal{G}$. Monotonicity, additivity and positive homogeneity from [F3] give ${\int }_{H_m^{+}}udP\ge {\int }_{H_m^{+}}vdP+P(H_m^{+})\mathrm{/}m\mathrm{.}$ Equality of the two event integrals forces $P(H_m^{+})=0$. The same argument with $u,v$ interchanged shows that $\{v\ge u+1\mathrm{/}m\}$ is null. Their countable union is the measurable discrepancy set $N_A=\{u\mathrm{\ne }v\}$, so $N_A$ is null. This proves the needed scalar uniqueness locally, without importing conditional-expectation uniqueness.

The set $N={\bigcup }_{A\in \mathcal{A}}{N}_A$ is in $\mathcal{G}$ and null by countability and [F5]. For any $\omega \mathrm{\notin }N$ the two probability measures agree on every member of $\mathcal{A}$. Their total masses are finite, so the locally supplied determination assertion [F2] yields equality on the entire target sigma-algebra. This gives a single exceptional set independent of the target event.

For $f={\mathbf{1}}_A$, the assertion is [F1]. If $f={\sum }_{j=1}^m{a}_j{\mathbf{1}}_{A_j}$ is nonnegative simple with disjoint measurable $A_j$ and finite coefficients, then $I_f={\sum }_j{a}_{j}K(\cdot ,A_j)$, so finite additivity of integrals proves the identity by summing the indicator identities. It includes the empty sum, giving $I_0=0$, and $f=1$, giving $I_1=1$.

For general nonnegative $f$, choose the prescribed simple approximants $s_n\uparrow f$ of [F4]. For every $\omega$, [F3] for the probability $K(\omega ,\cdot )$ gives $I_{s_n}(\omega )\uparrow I_f(\omega )$. The function $(\omega ,x)\mapsto f(x)$ is product-measurable since inverse images are $\mathrm{\Omega }\times f^{-1}(B)$; thus [F2] ensures $\mathcal{G}$-measurability of $I_f$. Applying [F3] on each $H$ on both sides of step 1.1 proves the displayed identity, allowing infinity. No representative selection or AC occurred. If using conditional-class notation, [F5] identifies this characterized nonnegative function with the class, under [F7].

For real $f$ with $C=E\mathrm{\mid }f(X)\mathrm{\mid }<\mathrm{\infty }$, step 2.1 at $\mathrm{\mid }f\mathrm{\mid }$ and $H=\mathrm{\Omega }$ gives $\int I_{\mathrm{\mid }f\mathrm{\mid }}dP=C$. Also $C=\int \mathrm{\mid }f\mathrm{\mid }d{P}_X$ by [F8], checking the equivalent law integrability condition. The set $D=\{I_{\mathrm{\mid }f\mathrm{\mid }}<\mathrm{\infty }\}$ is measurable. For every positive integer $n$, $nP(D^c)\le \int I_{\mathrm{\mid }f\mathrm{\mid }}=C$, hence $P(D^c)=0$. On $D$, both $I_{f^{+}},I_{f^{-}}$ are finite and their difference $J$ is the signed integral. Fill $J$ by zero off $D$; it is real measurable by [F2], and $\mathrm{\mid }J\mathrm{\mid }\le I_{\mathrm{\mid }f\mathrm{\mid }}$ on $D$, so $J$ is integrable. The two nonnegative identities from step 2.1 have finite integrals, at most $C$, so subtraction gives ${\int }_{H}JdP={\int }_{H}f(X)dP$. Removing $D^c$ changes neither integral, since it is measurable null. Thus $J$ is a conditional version; [F6] gives its almost-sure uniqueness, and [F7] supplies only the inherited class notation. No infinity is subtracted from infinity.

Fix $e_0\in E$ and the coding $c:E\to B\subseteq [0,1]$ supplied by [F2]. Write $A_q=c^{-1}[B\cap (-\mathrm{\infty },q]]$ and $u_q(\omega )=L(\omega ,A_q)$. The family $\{Y^{-1}(D):D\in \mathcal{T}\}$ is already a sigma-algebra and equals $\sigma (Y)$. For each rational q, integer $n\ge 1$ and $0\le k<2^n$, lift the event $\{k{2}^{-n}\le u_q<(k+1)2^{-n}\}$ to $D_{q,n,k}\in \mathcal{T}$, and lift $\{u_q=1\}$ to $D_{q,n,2^n}$. Choose this countable family using [F7]–[F9]. Set $h_{q,n}(y)=\min \text{\hspace{0.17em}⁣}(1,{\sum }_{k=0}^{2^n}k{2}^{-n}{\mathbf{1}}_{D_{q,n,k}}(y)),h_q(y)={\mathrm{lim\hspace{0.17em}sup}}_n{h}_{q,n}(y)\mathrm{.}$ By [F6] these functions are measurable. On the image of Y the lifted bins are disjoint, so $h_{q,n}(Y(\omega ))$ is the dyadic lower approximation to $u_q(\omega )$, exact at one. Hence $h_q\circ Y=u_q$ everywhere.

Let $T_0$ consist of y for which all rational comparisons $h_q\le h_r$ for $q<r$, tails $h_n\to 1$, $h_{-n}\to 0$, and right limits $h_{q+1\mathrm{/}n}\to h_q$ hold. By [F6]–[F8], this is measurable. For every $\omega$, the pushforward ${\eta }_{\omega }(C)=L(\omega ,c^{-1}[B\cap C])$ is a Borel probability whose rational CDF values are $u_q(\omega )$. The forward CDF clause of [F3] gives all the tests, so $Y(\omega )\in T_0$. Replace $h_q$ off $T_0$ by ${\mathbf{1}}_{\{q\ge 0\}}$ and call the result ${\widehat{h}}_q$.

For $y\in T$, define $F_y(x)={\inf }_{q\in \mathbb{Q},q>x}{\widehat{h}}_q(y)$. As in the real-kernel construction, it is nondecreasing, has tails zero and one, agrees with ${\widehat{h}}_r$ at every rational r, and is right-continuous. The existence clause of [F3], under [F9], gives a Borel probability ${\nu }_y$ with this CDF. It is unique locally: the equality class of two candidates is a lambda-system containing the closed-half-line pi-system, so [F4] and [F5] give equality on all Borel sets. Hence no family choice is needed. Fixed half-line evaluations are measurable by [F6]. The class of Borel C with measurable $y\mapsto {\nu }_y(C)$ contains those half-lines and the whole line, is closed under complements, and is closed under disjoint unions because sectionwise countable additivity expresses the evaluation as a measurable pointwise limit of partial sums. Thus [F4]–[F6] give all Borel C, producing an everywhere real probability kernel without a measure on T.

At $y=Y(\omega )$, the probabilities ${\nu }_y$ and ${\eta }_{\omega }$ agree at every rational closed cut. Their equality class is a lambda-system, and those cuts form a pi-system generating the Borel sets by F5, so F4 gives ${\nu }_{Y(\omega )}={\eta }_{\omega }$ directly. Thus the measurable set $T_1=\{y:{\nu }_y(B)=1\}$ contains the full image of Y. Put $K(y,A)=\{\begin{array}{ll}{\nu }_y(c[A]),& y\in T_1,\\ {\mathbf{1}}_A(e_0),& y\mathrm{\notin }{T}_1\mathrm{.}\end{array}$ Bimeasurability makes each evaluation measurable; support on B and the Dirac filler make every section a probability. For every $\omega ,A$, $K(Y(\omega ),A)=L(\omega ,A)$, so the exceptional set is empty and substitution in L's identities proves [F1].