Let $(X, {\cal E}, \mu)$ be a measure space and let $B \in {\cal E}$. Define $\mu_B:{\cal E} \to [0,\infty]$ by $\mu_B (E) = \mu(E\cap B)$.

Question

Let $(X, {\cal E}, \mu)$ be a measure space and let $B \in {\cal E}$. Define $\mu_B:{\cal E} \to [0,\infty]$ by $\mu_B (E) = \mu(E\cap B)$. (I have already proved that this does indeed define a new measure).

Find a necessary and sufficient condition on sets $B,C \in \mathcal{E}$ such that $\mu_B \leq \mu_C$.

So far, I have found that $B\subseteq C$ is a sufficient, but not necessary condition. Any ideas?

Answer 1

The condition is $\mu (B\setminus C)=0$. If this condition holds the $\mu (E\cap B) \leq \mu (E\cap C)+\mu (B\setminus C)=\mu (E\cap C)$ for all $E$. Conversely, if the inequality holds for all $E$ take $E=C^{c}$ to get $\mu (B\setminus C)=0$.