Let $(X, \mathcal{A}, \mu)$ be a measure space and $A,B \in \mathcal{A}$ with $0<\mu(A)<\infty$.
How to compute the average of the indicator function $\mathbf1_{B}$,
$⨍_A \mathbf1_{B}\;d\mu$ in terms of $\mu$?
Since the definition of the average of an integral is $⨍_A f=⨍_A f \ d\mu:=\frac{1}{\mu(A)} \cdot \int_A f \ d \mu$ I tried:
$⨍_A \mathbf1_{B}\;d\mu = \frac{1}{\mu(A)} \cdot \int_A \mathbf1_{B} \ d \mu$.
Here I'm not sure how to continue as $\mu(A)$ isn't given.
I thought about using:
$⨍_A \mathbf1_{B}\;d\mu = ⨍_X \mathbf1_{B} \cdot \mathbf1_{B}\;d\mu$, but it doesn't work.