I've been studying some statistics by the ways of measure theory, and came up with a problem in understanding conditional probability. The book gives the following definitions:

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G} \subset \mathcal{F}$ be a $\sigma$-algebra and $A \in \mathcal{F}$. Then, we define a probability measure on $\mathcal{G}$ by the law $$\nu(G) = \mathbb{P}(A \cap G).$$ If $\mathbb{P}(G) = 0$, then $$\nu(G) = \mathbb{P}(G \cap A) \leq \mathbb{P}(G) = 0,$$ hence $\nu \ll \mathbb{P}$. We define $\mathbb{P}(A \mid \mathcal{G})$ as $d\nu/d\mathbb{P}$. This way, the conditional probability is kind of well-defined, since Radon-Nikodym derivatives are equal almost everywhere.

My problem here is: I don't understand what is the intuition behind this definition. For some help in understanding my problem, there is an important result

The function $\mathbb{P}(A \mid \mathcal{G})$ has the following properties almost everywhere:

  • $0 \leq \mathbb{P}(A \mid \mathcal{G}) \leq 1$;
  • $\mathbb{P}(\emptyset \mid \mathcal{G}) = 0$;
  • If $\{A_n\}_{n = 1}^\infty$ is a sequence of two-by-two disjoint sets in $\mathcal{F}$, then $$\mathbb{P}\left(\bigcup_{n=1}^\infty A_n \mid \mathcal{G}\right) = \sum_{n = 1}^\infty \mathbb{P}(A_n \mid \mathcal{G}).$$

This lemma says that, for almost every point, $\mathbb{P}(A \mid \mathcal{G})$ is a probability measure in $\mathcal{F}$ (as a function of $A$). Now, I don't have a single clue about what a "field" of probability measures mean and how it is related in any way with concepts of elementary probability. Even more, is there any "real-world" applications of this concept? With real-world applications I mean like some kind of elementary statistics problem that can be solved using this, some applicable situation that might appear in a non-math context.

  • $\begingroup$ Could you please explain what "two-by-two disjoint" means and how it differs from "disjoint"? And what does it mean for a set function to be a probability measure at "almost every point"? For intuition on conditional expectations, please search our site. $\endgroup$
    – whuber
    Dec 27, 2022 at 14:28