Almost-Everywhere Fiber Support and Null-Set Versions

Why Fiber Support Is an Almost-Sure Statement

A measurable observation usually does not identify the original random element uniquely. If the observed value is $y$, the possible underlying values lie in the fiber over $y$. A conditional law given that observation should therefore assign probability one to that fiber—at least for almost every observation under its actual distribution.

The qualification “almost every” matters. Conditional laws are determined by integral identities, and those identities do not constrain their values at conditioning points that occur with probability zero. We will prove the fiber statement and then see explicitly how a valid version can fail it at such a point.

Conditional Laws Given a Map

Let $E$ and $F$ be standard Borel spaces, let $Z$ be an $E$-valued random element on a probability space with probability measure $\mathbb P$, and let $T:E\to F$ be measurable. Write $\mu$ for the law of $Z$, and $\nu$ for the law of $T(Z)$.

Definition: A regular conditional law of $Z$ given $T(Z)$ is a probability kernel $K$ from $F$ to $E$ such that, for every measurable $A\subseteq E$ and $C\subseteq F$, $$ \mathbb P\bigl(Z\in A,\ T(Z)\in C\bigr) = \int_C K(y,A)\,\nu(dy). $$ Here a probability kernel means that $K(y,\cdot)$ is a probability measure for each $y$, and $y\mapsto K(y,A)$ is measurable for each measurable $A$.

For $y\in F$, the fiber of $T$ over $y$ is $T^{-1}(\{y\})=\{x\in E:T(x)=y\}$. Saying that $K(y,\cdot)$ is concentrated on this fiber means precisely that $K(y,T^{-1}(\{y\}))=1$. This is a statement about probability assigned to the fiber; it does not assert that every point in the fiber has positive probability or belongs to the topological support.

The Almost-Everywhere Fiber Theorem

Theorem: If $E$ and $F$ are standard Borel spaces and $K$ is a regular conditional law of $Z$ given $T(Z)$, then $$ K\bigl(y,T^{-1}(\{y\})\bigr)=1 \quad\text{for }\nu\text{-almost every }y. $$

Proof. For each measurable $B\subseteq F$, define $$ L(y,B)=K\bigl(y,T^{-1}(B)\bigr). $$ For fixed $y$, $L(y,\cdot)$ is the probability measure obtained by mapping the measure $K(y,\cdot)$ through $T$. The defining identity for $K$, applied to $A=T^{-1}(B)$, gives, for every measurable $C\subseteq F$, $$ \int_C L(y,B)\,\nu(dy) = \mathbb P\bigl(T(Z)\in B,\ T(Z)\in C\bigr) = \nu(B\cap C) = \int_C \mathbf{1}_B(y)\,\nu(dy). $$ Consequently, for each fixed $B$, $L(y,B)=\mathbf{1}_B(y)$ for $\nu$-almost every $y$. Indeed, two bounded measurable functions whose integrals agree over every measurable $C$ agree almost everywhere.

Because $F$ is standard Borel, its measurable sets have a countable generating $\pi$-system $\mathcal{G}$, which we take to contain $F$. For each $B\in\mathcal{G}$, the equality just obtained can fail only on a $\nu$-null set. The union of these exceptional sets is still null because $\mathcal{G}$ is countable. Thus there is one $\nu$-full set $F_0$ such that, for every $y\in F_0$ and every $B\in\mathcal{G}$, $$ L(y,B)=\mathbf{1}_B(y)=\delta_y(B). $$ For each such $y$, both $L(y,\cdot)$ and the point mass $\delta_y$ are probability measures. Since they agree on a generating $\pi$-system, uniqueness of measures implies $L(y,\cdot)=\delta_y$ on all measurable subsets of $F$. In particular, evaluating this equality on the singleton $\{y\}$ gives $$ K\bigl(y,T^{-1}(\{y\})\bigr) = L(y,\{y\}) = \delta_y(\{y\}) = 1. $$ This holds for every $y\in F_0$, a set of full $\nu$-measure. $\square$

The standard Borel assumption provides the countable generating $\pi$-system used to pass from separate almost-everywhere equalities to a single full-measure set on which the probability measures agree. The conclusion is about the values of the kernel on that full-measure set; it does not say that every $y\in F$ must satisfy the fiber identity.

Worked Example: A Discrete Observation

Worked Example: Conditioning on the Sum of Two Bits

Let $Z=(U,V)$ be uniform on $\{(0,0),(0,1),(1,0),(1,1)\}$, and let $T(u,v)=u+v$. The possible observations are $0,1,2$. Define the conditional law by $$ K(0,\cdot)=\delta_{(0,0)},\qquad K(1,\cdot)=\tfrac12\delta_{(0,1)}+\tfrac12\delta_{(1,0)},\qquad K(2,\cdot)=\delta_{(1,1)}. $$ The observation probabilities are $\nu(\{0\})=1/4$, $\nu(\{1\})=1/2$, and $\nu(\{2\})=1/4$. For observation $0$, the only compatible point is $(0,0)$; for observation $1$, the two compatible points are $(0,1)$ and $(1,0)$; and for observation $2$, the only compatible point is $(1,1)$. Each stated conditional measure therefore assigns probability one to the appropriate fiber.

For example, if $A=\{(0,1)\}$ and $C=\{1\}$, then the left side of the conditional-law identity is $\mathbb P(Z\in A,T(Z)\in C)=1/4$. The right side is $$ \int_C K(y,A)\,\nu(dy) = K(1,A)\nu(\{1\}) = \tfrac12\cdot\tfrac12 = \tfrac14. $$ The same identity follows for any $A$ and $C$ by summing the probabilities of the finitely many points, so $K$ is indeed a conditional law.

Worked Example: A Non-Injective Map on an Interval

Worked Example: Conditioning on the Square

Let $X$ be uniform on $[-1,1]$ and set $T(x)=x^2$. The observation $Y=X^2$ lies in $[0,1]$. For $0<y<1$, the two points in the fiber are $\sqrt y$ and $-\sqrt y$. A conditional law is $$ K(y,\cdot)=\tfrac12\delta_{\sqrt y}+\tfrac12\delta_{-\sqrt y} \quad (0<y<1). $$ At $y=0$, set $K(0,\cdot)=\delta_0$, and at $y=1$, set $K(1,\cdot)=\tfrac12\delta_1+\tfrac12\delta_{-1}$. For $y\notin[0,1]$, set $K(y,\cdot)=\delta_0$; those values will not affect the conditional-law identity because $\nu$ assigns them probability zero.

To verify the formula, for Borel sets $A\subseteq\mathbb R$ and $C\subseteq[0,1]$, change variables separately on the positive and negative halves of the interval: $$ \mathbb P(X\in A,\ X^2\in C) = \int_0^1 \mathbf{1}_C(y) \frac{\mathbf{1}_A(\sqrt y)+\mathbf{1}_A(-\sqrt y)}{4\sqrt y}\,dy. $$ The law $\nu$ of $Y$ has density $1/(2\sqrt y)$ on $0<y<1$. Therefore integrating $K(y,A)$ against $\nu(dy)$ gives exactly the displayed expression. The endpoint choices do not affect this integral. For every $y\in(0,1)$, each point receiving mass lies in the fiber $x^2=y$, and the endpoint choices are also concentrated on their fibers.

Why Pointwise Fiber Support Can Fail

The disintegration identity integrates $K(y,A)$ with respect to $\nu(dy)$. Changing the kernel on a $\nu$-null set leaves every such integral unchanged. The following result makes this freedom precise.

Proposition: Let $N\subseteq F$ be measurable with $\nu(N)=0$, let $Q$ be any probability measure on $E$, and define $$ K'(y,A)= \begin{cases} Q(A),&y\in N,\\ K(y,A),&y\notin N. \end{cases} $$ Then $K'$ is also a regular conditional law of $Z$ given $T(Z)$.

Proof. For each measurable $A\subseteq E$, the function $y\mapsto K'(y,A)$ is measurable because it is defined by measurable functions on the measurable sets $N$ and $N^c$. For every $y$, $K'(y,\cdot)$ is a probability measure, either $Q$ or $K(y,\cdot)$, so $K'$ is a probability kernel. Finally, for measurable $C\subseteq F$, $$ \int_C K'(y,A)\,\nu(dy) = \int_{C\setminus N}K(y,A)\,\nu(dy) + \int_{C\cap N}Q(A)\,\nu(dy) = \int_C K(y,A)\,\nu(dy), $$ because $\nu(C\cap N)=0$. The final expression is the required joint probability by the defining identity for $K$. Hence $K'$ is another version of the same conditional law. $\square$

Worked Example: Altering an Identity-Map Conditional Law

Let $X$ be uniform on $[0,1]$ and take $T(x)=x$. The kernel $K(y,\cdot)=\delta_y$ is a conditional law: for Borel $A,C\subseteq\mathbb R$, $$ \int_C K(y,A)\,\nu(dy) = \int_C \mathbf{1}_A(y)\,\nu(dy) = \mathbb P(X\in A,\ X\in C). $$ Now change only its value at $y=0$, setting $K'(0,\cdot)=\delta_1$ and $K'(y,\cdot)=\delta_y$ for $y\ne0$. Since $\nu(\{0\})=0$, the proposition shows that $K'$ is still a conditional law. Yet the fiber over $0$ is $T^{-1}(\{0\})=\{0\}$, and $$ K'(0,T^{-1}(\{0\}))=\delta_1(\{0\})=0. $$ Thus a valid version can fail fiber concentration at an individual null conditioning value.

How to Read the Conclusion

The theorem does not say that a conditional law “given $T(Z)=y$” is uniquely determined for every $y$. The precise conclusion is that every version has the fiber property for $\nu$-almost every $y$. Two versions can disagree at exceptional values while representing the same conditional law.

This distinction is useful whenever conditional distributions are evaluated at a specific observation, especially when that observation has probability zero. A formula may provide a natural or continuous choice at such a point, but that choice comes from the formula or additional structure—not from the defining integral identity alone. In applications, it is therefore important to distinguish a pointwise formula for a chosen version from a statement that follows for all versions almost everywhere.

A common pitfall is to infer that the theorem requires $K(y,\cdot)$ to live on the fiber for every $y$, including values outside the range of $T$ or values with $\nu$-probability zero. The proof only establishes equality outside one $\nu$-null set. On that set, the kernel may be altered without changing the conditional law.

Check Your Understanding

Use the definitions and proof above to answer the following questions.

What does it mean for $K(y,\cdot)$ to be concentrated on the fiber of $T$ over $y$?
In the theorem’s proof, why is a countable generating $\pi$-system useful?
Why does equality of the integral identities remain true after changing a kernel on a $\nu$-null set?
For the square map $T(x)=x^2$, which points receive conditional mass when $0<y<1$, and why are they compatible with the observation?
In the identity-map example, why is the modified kernel still valid even though it fails the fiber property at $y=0$?