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There are perhaps some related questions on this site already (I can think of this question, this or this), but I want to be more concrete here. In the linked questions, and e.g. in Dudley's book (Real analysis and probability), they claim the following concerning the fact that a conditional probability $\mathbb{P}(A\mid \mathcal{C})(\omega)$ can fail to be a measure (as a function of $A$) for all $\omega$:

... countable additivity will hold for almost all $\omega$ ... But the set of zero probability where countable additivity fails might depend on the sequence, and the union of all such sets might cover $\Omega$.

The quote is from Dudley's book, but the links say something very similar. Reading this sentence over and over again, I still don't quite understand why we need the concept of regular conditional probability. What is an example of a probability space where to each sequence of disjoint sets, we have a null set and that the union of the null sets has positive probability?

I was thinking about $([0,1],\mathcal{B}([0,1]))$ and Lebesgue measure, for which $\{x\}$ is a null set and whose union over all $x\in[0,1]$ is $[0,1]$, a non-null set. But I can't connect $\{x\}$ to a sequence of disjoint sets where countable additivity fails.

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Given a sequence $A_i, i \in \mathbb N$ of disjoint subsets of $[0,1]$, we can use the Axiom of Choice to construct a set $N=\{a_i | a_i \in A_i, i \in \mathbb N\}$. Clearly $N$ is a null set because it is the countable union of singletons which are null sets. Given any element $x \in [0,1]$, we can construct a sequence of disjoint sets $A^x_i, i \in \mathbb N$ where $A^x_0=\{x\}$ (again using Axiom of Choice). Consider the null set $N^x$ constructed from the sequence $A^x$ by the above method ($N^x=\{a^x_i | a^x_i \in A^x_i, i \in \mathbb N\}$). Since the only possible choice for $a^x_0$ is $x$, it is clear that $x \in N^x$. The union $\cup_{x\in [0,1]} N^x = [0,1]$ which is a non null set.

Avoiding Axiom of Choice

On second thought, we can get rid of the Axiom of Choice by using singleton sets. The axiom is not needed in the first place where I used it if each $A_i$ has only one element. In fact in this case, $N=\cup A_i$.

In the second application of the Axiom, we can let $A^x_i= \{x^{i+1}\}$ for $x \in (0,1)$. The union $\cup_{x\in (0,1)} N^x = (0,1)$ which is a non null set.

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  • $\begingroup$ Can you explain how we construct a sequence of disjoint sets $A_i^x,i\in\mathbb{N}$ given $x$ (and possibly using axiom of choice)? $\endgroup$ Commented Dec 3 at 16:14
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    $\begingroup$ Have edited the answer to provide an explicit construction eliminating the use of the Axiom of Choice completely. $\endgroup$ Commented Dec 4 at 5:44
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Let $(\Omega,\mathcal{A},P)$ be a probability space and $\mathcal{C}\subseteq\mathcal{A}$ a sub-$\sigma$-algebra. For $A\in\mathcal{A}$, a (!) conditional probability of $A$ given $\mathcal{C}$ is a $\mathcal{C}$-measurable function $Y_A:\Omega\to\mathbb{R}$ such that $P(A\cap C)=\int_C Y_A~\mathrm{d}P$ for all $C\in\mathcal{C}$. One can restrict oneself to conditional probabilities with values in $[0,1]$, and we will do so. Importantly, if $Y_A'$ is $\mathcal{C}$-measurable and $P$-almost surely equal to $Y_A$, then $Y'_A$ is also a conditional probability of $A$ given $\mathcal{C}$.

A regular conditional probability with respect to $\mathcal{C}$ is a probability kernel $\kappa:\Omega\times\mathcal{A}\to[0,1]$ if $\kappa(\cdot,A)$ is a conditional probability of $A$ given $\mathcal{C}$ for all $A$.

So for a regular conditional probability, for each $A$, a specific conditional probability is chosen. The question is what happens if we go the other way. So we choose for each $A$ a conditional probability $Y_A$. We then define $\kappa:\Omega\times\mathcal{A}\to[0,1]$ by $\kappa(x,A)=Y_A(x)$. The question is whether this will automatically define a regular conditional probability. In general, the answer is no. There are examples where this works for no choice of the $Y_A$ (see Exercise 6 in the relevant chapter of Dudley's book). But there are also cases where some choice does not work even though others would. Here is an eample:

Let $\Omega=[0,1]^2$, $\mathcal{A}$ be the Borel sets, $P=\lambda\otimes\lambda$ be the uniform distribution, and $\mathcal{C}$ be the $\sigma$-algebra generated by the first coordinate projection. For all $A\in\mathcal{A}$, let $Y_A(x,y)=\lambda(A_x)$ with $A_x=\{y:(x,y)\in A\}$. Then $Y_A$ is a regular conditional probability of $A$ with respect to $\mathcal{C}$. Unless $A$ is of the form $A=\{(z,z)\}$, let $Y'_A=Y_A$. And for each $z$, let $Y_{\{(z,z)\}}'$ be given by $Y_{\{(z,z)\}}'(x,y)=1$ if $x=z$ and $Y_{\{(z,z)\}}'(x,y)=0$ otherwise. Then, each $Y'_A$ is a regular conditional probability of $A$ with respect to $\mathcal{C}$. However, if we define $\kappa$ by $\kappa((x,y),A)=Y_A'(x,y)$, then $\kappa((x,y),\cdot)$ is never a probability measure, because $\kappa((x,y),\{(x,x)\})=1$ but $\kappa((x,y),\{(x,x),(x,0),(x,1)\})=0$.

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    $\begingroup$ look who's here $\endgroup$ Commented Dec 3 at 16:03

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