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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.

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    $\begingroup$ It gives $\frac{\mu(A\cap B)}{\mu(A)}.$ You can't do better without more information $\endgroup$ – idm Dec 15 '18 at 14:35
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    $\begingroup$ $\int_A \mathbf1_{B} \ d \mu=\mu(A\cap B)$ $\endgroup$ – Matematleta Dec 15 '18 at 14:36

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