Understanding what a probability measure can be hard, or at least feel harder than it should be — it’s just a function after all, right? Yes, but it’s a function which takes a set as an input, and that’s what makes it hard or at least initially unintuitive to think about. Probability measures are important to understand, though, so we’ve adopted. We think of them as functions which take in a set and return the corresponding probability associated with that set.
$$ \mathbb{P}: \mathcal{F} \to [0,1] $$
In many regards, probabilities measures drive everything we’ve talked about so far. They are related to the sample space, they’re the basis of expected values. They’re central. But it turns out that there’s a notion — a conditional probability measure — which is almost even more fundamental to what’s going on — and arguably harder to wrap one’s mind around. Let’s define its type signature and then explain what’s going on.
$$ \mu : \Sigma(X) \to \Omega \to (\mathcal{F} \to [0,1]) $$
Now, as we known, brackets don’t matter in a type signature, but they’re can make it easier to understand. This says that the conditional probability measure takes in a $\sigma$-algebra, an element of the sample space and returns a probability measure. In practice, we’ll often see if defined as follows, where the $\sigma$-algebra is fixed:
$$ \mu_{\sigma(X)}(\cdot, \cdot) $$
with the condition that:
$$ \begin{align*}\forall B \in \sigma(X), \forall A \in \mathcal{F}, \int {B} \mu{\sigma(X)}(\cdot, A)d\mathbb{P} &= \mathbb{P}(A \cap B) \end{align*} $$
$$ \forall A \in \sigma(\tilde{Y}(1)), \forall B \in \sigma(D),\quad \mu_{\sigma(X)}(\cdot, A\cap B) = \mu_{\sigma(X)}(\cdot,A)\mu_{\sigma(X)}(\cdot, B) $$
Does the following hold?
$$ \int {B} \mu{\sigma(X)}(\cdot, A) = \mathbb{P}(B \cap A) \iff \int _B\mathbb{E}[Y \vert X] d\mathbb{P} = \int _B X d\mathbb{P} $$