Let $\mu$ be an outer measure on $X$ and $B \subset X$. We define the restriction of $\mu$ to $B$.
$\mu\big|_{B}:\mathcal{P}(X) \to [0,\infty]$, by $\mu\big|_{B}=\mu(B\cap A).$
Show that:
If $B$ is $\mu$-measurable and $A\subset B$ is too, then $A$ is $\mu\big|_{B}$-measurable.
We know that for $C \subset X, \mu(C) = \mu(C \cap B)+\mu(C\cap B^c)$. I have to show that $$\mu\big|_{B}(C) \ge \mu\big|_{B}(C\cap A)+\mu\big|_{B}(C\cap A^c).$$ If I expand $\mu\big|_{B}(C)$ out, $\mu\big|_{B}(C)= \mu(B\cap C)$. I think that I have to use the fact that $\mu(C) = \mu(C \cap B)+\mu(C\cap B^c)$, but don't know how to do it. Could you give some help?
Thank you in advance.