Given the measure space $(\Omega, A, P)$ and $X:\Omega \rightarrow \mathbb{R}$ an integrable function. D is a closed set of $\mathbb{R}$ such that $\forall B\in A$ with $P(B)>0 $
$\displaystyle\frac{1}{P(B)}\int_{B}XdP \in D$ . Prove that $P\{X\in D\}=1$.
this looks quite similar to the intermediate value for integrals taking the riemann integral in the sigma algebra of the borelians. so i tried to get an upper bound and a lower bound for X such that when i integrate i have like :
$P(B)m\leq\int_{B}XdP \leq M P(B)$ then i divide and that's it but what is there any thing that asures me that i can get those bounds? got any other ideas?