# Average (determine with Lebesgue integral/measure)

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.

$$⨍_A \mathbf1_{B}\;d\mu = ⨍_X \mathbf1_{B} \cdot \mathbf1_{B}\;d\mu$$, but it doesn't work.
• It gives $\frac{\mu(A\cap B)}{\mu(A)}.$ You can't do better without more information – idm Dec 15 '18 at 14:35
• $\int_A \mathbf1_{B} \ d \mu=\mu(A\cap B)$ – Matematleta Dec 15 '18 at 14:36