Suppose $(X,M,\mu)$ is a measure space and there exists a set $A\in M$ with $\mu(A)<\infty$ such that $\mu(E\cap A)=\mu(E) \mu(A)$ of all $E\in M$.
I want to show that $\int_A f \, d\mu =\mu(A) \int_X f \, d\mu$. for all $f \in L^1$
I know the condition $\mu(E\cap A)=\mu(E) \mu(A)$ will tell us the space is finite. and I was only able to prove that its true for simple function that is $f=\chi_E$.
Update:- this solution was based on the comments of Michael
$$\int_X f \, d\mu =\int_A f \, d\mu+ \int_{A^c} f d\mu$$ we have two case:-
- When $\mu(A)=0$ and $\mu(A^c)=\mu(X)$ then:-
$$\int_A f d\mu=0=0\times \int_{X} f d\mu$$
- When $\mu(A)=1=\mu(X)$ and $\mu(A^c)=0$ then:- $$ 1\times \int_X f \, d\mu =\int_A f \, d\mu+0$$
Is this right now? Note:- this solution depend on the fact that integral over null set is zero.