Question about an outer measure of a real subset

Question

Let $B$ be a subset of the real numbers, and let $\mu$ be an outer measure, and let $A$ be the union of a finite number of real intervals. I have to show that $\mu(B)=\mu(B\cup A)+\mu(B\cap A^c)$, where $A^c$ is the complement of $A$ with respect to the real numbers. I know that $$B=(B\cap A)\cup (B\cap A^c)$$ and so, by subadditivity of outer measure, $$\mu(B)\le\mu(B\cup A)+\mu(B\cap A^c)$$ I need help proving $$\mu(B)\ge\mu(B\cup A)+\mu(B\cap A^c)$$

Answer 1

It is not true. Consider the outer measure $$ \mu(E)=\begin{cases} 1,&\ E\text{ is uncountable}\\0,&\ E\text{ is countable}\end{cases} $$ Then, with $B=[0,1]$, $A=[0,2]$, you have $$ \mu(B)=1 <2=\mu(B\cap A)+\mu(B\cap A^c). $$