Why Existence Is a Separate Question
A conditional expectation given a random element can be represented by a measurable function of that element. A conditional distribution asks for more: it seeks one probability measure for each possible value of the conditioning element, with those measures varying measurably. The exceptional null set in a conditional expectation may depend on the event being conditioned, so choosing versions separately does not by itself give a probability measure at every conditioning value.
Standard Borel spaces provide the structure needed to make these choices consistent. The proof below reduces the target space to a Borel subset of the real line, constructs a conditional distribution function using countably many Radon–Nikodym derivatives, and then applies measure extension.
This identity says that \(K(Y,A)\) is a version of the conditional probability of \(\{X\in A\}\) given the information generated by \(Y\). The definition requires the kernel to be a probability measure at every \(y\), not merely almost surely.
Existence Theorem
The conditioning space \(F\) matters because the kernel must be a measurable function of the conditioning value. The target space \(E\) matters in the construction: its standard Borel structure lets us encode its measurable sets using real thresholds and recover probability measures by a one-dimensional distribution function.
Worked Example: A Deterministic Conditional Law
Worked Example: A Deterministic Conditional Law
Let \(Y\) be real-valued and let \(X=Y^2\). Define \(K(y,A)=\mathbf 1_A(y^2)\) for each Borel set \(A\subseteq\mathbb R\). For fixed \(y\), this is the point mass at \(y^2\), hence a probability measure; for fixed \(A\), it is measurable in \(y\), since \(y\mapsto y^2\) is measurable.
For every Borel set \(C\subseteq\mathbb R\), the event \(\{X\in A,Y\in C\}\) is \(\{Y^2\in A,Y\in C\}\). Also \(K(Y,A)=\mathbf 1_A(Y^2)=\mathbf 1_{\{X\in A\}}\). Therefore $$ \int_{\{Y\in C\}}K(Y,A)\,d\mathbb P = \mathbb P(X\in A,Y\in C), $$ so \(K\) is a regular conditional distribution.
Proof by Countable Reduction
We prove the theorem. A standard Borel space is measurably isomorphic to a Borel subset \(B\) of \([0,1]\). Thus there is a measurable bijection \(\phi:E\to B\) whose inverse is measurable. Write \(Z=\phi(X)\), viewed as a real-valued random element. We first construct a conditional distribution for \(Z\), then transfer it back to \(E\).
Let \(\nu\) be the distribution of \(Y\) on \(F\). For each rational number \(q\), define a finite measure on \(F\) by $$ \nu_q(C)=\mathbb P(Z\le q,\ Y\in C),\qquad C\in\mathcal F. $$ Since \(0\le \nu_q(C)\le \nu(C)\), the Radon–Nikodym theorem gives a measurable function \(f_q:F\to[0,1]\), defined up to a \(\nu\)-null set, such that $$ \nu_q(C)=\int_C f_q(y)\,\nu(dy). $$ There are only countably many rational numbers. We can therefore choose versions and remove one \(\nu\)-null set so that, outside it, \(f_q(y)\le f_r(y)\) whenever \(q\le r\). Indeed, each inequality holds almost everywhere because \(\nu_q\le\nu_r\), and there are only countably many such pairs. Choose \(f_q=0\) for \(q<0\) and \(f_q=1\) for \(q\ge1\), which are valid versions because \(Z\in B\subseteq[0,1]\) almost surely.
We also need right continuity at rational thresholds. For each rational \(q\), choose rational numbers \(r_n(q)>q\) decreasing to \(q\). The events \(\{Z\le r_n(q)\}\) decrease to \(\{Z\le q\}\). After imposing the monotonicity inequalities, the functions \(f_{r_n(q)}\) decrease pointwise outside the same null set. Their limit has the same integrals over every \(C\in\mathcal F\) as \(f_q\), by continuity of probability for decreasing events and monotone convergence applied to \(1-f_{r_n(q)}\). Hence that limit equals \(f_q\) almost everywhere. Since the rational \(q\)'s are countable, one more common null set suffices to ensure $$ f_q(y)=\inf_{\substack{r\in\mathbb Q\\r>q}} f_r(y) $$ for every rational \(q\) at every \(y\) outside that set. To see that the selected sequence gives this infimum, use monotonicity and the fact that for every rational \(r>q\), eventually \(r_n(q)<r\).
Call the combined exceptional set \(N\). For \(y\notin N\), define, for real \(x\), $$ F_y(x)=\inf_{\substack{q\in\mathbb Q\\q>x}} f_q(y). $$ This function is nondecreasing in \(x\). It is right-continuous: if \(x_n\downarrow x\), the sets of rationals \(\{q:q>x_n\}\) increase to \(\{q:q>x\}\), so the corresponding infima decrease to \(F_y(x)\). It equals zero for \(x<0\) and one for \(x\ge1\), by the chosen values of \(f_q\). Thus its limits at minus and plus infinity are zero and one.
Apply this fact to \(F_y\), obtaining a probability measure \(\mu_y\) for each \(y\notin N\). At rational \(q\), the construction and the right-continuity identity give \(F_y(q)=f_q(y)\). For \(y\in N\), temporarily choose any fixed probability measure on \(\mathbb R\), for example the point mass at zero. Thus \(\mu_y((-\infty,q])=f_q(y)\) outside \(N\).
Measurability and the Conditional-Law Identity
Fix \(C\in\mathcal F\). For rational \(q\), the Radon–Nikodym identity gives $$ \int_C\mu_y((-\infty,q])\,\nu(dy) = \mathbb P(Z\le q,\ Y\in C). $$ The class of Borel sets \(A\subseteq\mathbb R\) for which $$ \int_C\mu_y(A)\,\nu(dy)=\mathbb P(Z\in A,\ Y\in C) $$ holds is closed under complements and countable disjoint unions, and contains the rational half-lines. Another application of the pi-lambda theorem shows that the identity holds for every Borel \(A\). Since \(C\) was arbitrary, it holds for every \(C\in\mathcal F\).
The measure \(\mu_y\) is supported on \(B\) for almost every \(y\). In fact, the identity just proved, with \(A=B\) and \(C=F\), gives $$ \int_F\mu_y(B)\,\nu(dy)=\mathbb P(Z\in B)=1. $$ Because \(0\le\mu_y(B)\le1\), this implies \(\mu_y(B)=1\) for \(\nu\)-almost every \(y\). The set \(N'=\{y:\mu_y(B)<1\}\) is measurable and \(\nu\)-null. On \(N'\), replace \(\mu_y\) by the point mass at a fixed \(b_0\in B\). Such a point exists because \(Z\) takes values in \(B\). This change preserves measurability and all the integral identities, since \(N'\) is null, and now \(\mu_y(B)=1\) for every \(y\).
For \(A\in\mathcal E\), the set \(\phi(A)\) is Borel in \(B\), hence Borel in \(\mathbb R\). Define $$ K(y,A)=\mu_y(\phi(A)). $$ For each \(y\), this is a probability measure on \(E\), because \(\mu_y(B)=1\) and \(\phi\) is a measurable isomorphism. For each \(A\), it is measurable in \(y\). Applying the identity for \(\mu_y\) to \(\phi(A)\) gives $$ \int_{\{Y\in C\}}K(Y,A)\,d\mathbb P = \mathbb P(X\in A,\ Y\in C). $$ Thus \(K\) is a regular conditional distribution of \(X\) given \(Y\). \(\square\)
Worked Example: Conditioning on an Independent Element
Worked Example: Conditioning on an Independent Element
Suppose \(X\) and \(Y\) are independent, with \(X\) taking values in a standard Borel space \(E\). Set \(K(y,A)=\mathbb P(X\in A)\) for every \(y\) and \(A\in\mathcal E\). This is a probability kernel: it is a probability measure in \(A\), and it is constant, hence measurable, in \(y\). Independence gives $$ \int_{\{Y\in C\}}K(Y,A)\,d\mathbb P = \mathbb P(X\in A)\mathbb P(Y\in C) = \mathbb P(X\in A,\ Y\in C). $$ So the unconditional law of \(X\), used constantly across conditioning values, is a regular conditional distribution.
Worked Example: Finite Values and an Unattained Conditioning Value
Worked Example: Finite Values and an Unattained Conditioning Value
Let \(\Omega=\{1,2,3,4\}\) with equal probabilities, and set \(Y(1)=Y(2)=0\), \(Y(3)=1\), \(Y(4)=2\). Let \(X(1)=0\), \(X(2)=1\), \(X(3)=1\), and \(X(4)=0\). Define a kernel on \(F=\{0,1,2,3\}\) by $$ K(0,\cdot)=\tfrac12\delta_0+\tfrac12\delta_1,\qquad K(1,\cdot)=\delta_1,\qquad K(2,\cdot)=\delta_0,\qquad K(3,\cdot)=\delta_0. $$ Each value is a probability measure, and every function on this finite \(F\) is measurable. For each attained value of \(Y\), the assigned measure is exactly the distribution of \(X\) on the corresponding fiber: at \(Y=0\), the two equally likely outcomes give \(X=0\) and \(X=1\); at \(Y=1\), \(X=1\); and at \(Y=2\), \(X=0\). Summing over any subset \(C\subseteq F\) therefore gives the defining identity. The value at \(3\), which is never attained, may be chosen freely without changing that identity.
Why the Common Null Set Matters
The countability in the proof is not cosmetic. A Radon–Nikodym derivative for each event is defined only up to a null set. If one chooses versions independently for every target event, there may be no single exceptional set outside which all probability-measure identities hold at once. The rational half-lines reduce the initial choices to a countable family, so monotonicity and right continuity can be enforced outside one common null set. Measure extension then supplies the values on all Borel sets.
The theorem guarantees existence; the construction does not require the conditional law to have a density with respect to Lebesgue measure. Conditional laws can have atoms, be singular, or mix these behaviors. The next issue is how much freedom remains in the choice of a kernel on null sets; existence alone does not settle that question.
Check Your Understanding
Try to answer these questions using the definitions and construction above.
- Why does the Radon–Nikodym theorem apply to the measures \(\nu_q\) and \(\nu\)?
- Why can the monotonicity inequalities for rational thresholds be arranged outside one null set?
- What property of \(F_y\) permits the Lebesgue–Stieltjes measure extension?
- Why is it necessary to modify the constructed measures if they assign positive mass outside \(B\)?
- In the finite example, why does the kernel value at \(Y=3\) not affect the conditional-law identity?