I wonder about how to explain best why the following formula (1) is called "Bayes' formula".
Given a probability space $(\Omega,\mathfrak A,P)$, a sub-$\sigma$-algebra $\mathfrak B\subset \mathfrak A$, a strictly positive random variable $Z\in L^1(\mathbb P)$ with $E[Z] =1$, we can define the equivalent probability measure $Q$ via $Q(A) = E[Z \textbf 1_A]$. Then, for any $Y\in L^1(\Omega,\mathfrak A,Q)$, we have \begin{equation} (1) \qquad E^Q[Y \mid \mathfrak B] = \frac{E[YZ\mid \mathfrak B]}{E[Z \mid \mathfrak B]}. \end{equation} Note that, if $A \in \mathfrak A$ with $P(A) > 0$, then we can put $Z = \textbf 1_A / P(A)$ in order to get $Q = P(\cdot\mid A)$.
But what is the precise relation to the elementary formula due to Bayes/Laplace, i.e. for $A,B \in \mathfrak A$ with $P(A)P(B) >0$ we have \begin{equation*} P(B\mid A) = \frac{P(B)P(A\mid B)}{P(A)}, \end{equation*} how do we have to choose $\mathfrak A, \mathfrak B, Z,Y$ above in order to derive this formula (without using the elementary argument which shows the elementary rule)?
Otherwise, there would – in my humble opinion – be no reason to call it "Bayes' formula".
Let $\mathfrak B$ be trivial, and set $Y = 1_B$, $Z = 1_A / P(A)$. Then:
$$ P(B \mid A) = E^Q[Y \mid \mathfrak B] = \frac{E[YZ \mid \mathfrak B]}{E[Z \mid \mathfrak B]} = \frac{P(A \cap B)}{P(A)},$$
which isn't exactly Bayes' rule, but it's pretty close; you can of course derive Bayes' rule by applying this result twice.