showing that a subset of a $\mu^*$-measurable set is also $\mu^*$-measurable I am trying to prove the following:
Let $X$ be a set, $\mu^*$ an outer measure on $X$, and $Y$ a $\mu^*$-measurable subset of $X$ such that $\mu^*(Y)<\infty$. Let $B$ be a subset of  $Y$ such that $\mu^*(Y)=\mu^*(B)+\mu^*(Y\cap B^c)$. Prove that $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ for every $\mu^*$-measurable subset $A$ of $X$.
Here is what I have so far:
Since $Y$ is $\mu^*$-measurable, then we know that
$$\mu^*(A)=\mu^*(A\cap Y)+\mu^*(A\cap Y^c)=\mu^*(A\cap(B\cup (Y\cap B^c)))+\mu^*(A\cap Y^c)=\mu^*((A\cap B)\cup (A\cap(Y\cup B^c)))+\mu^*(A\cap Y^c).$$
From here I wanted to use the disjointedness of $A\cap B$ and $A\cap(Y\cup B^c)$ to write $\mu^*((A\cap B)\cup (A\cap(Y\cup B^c)))=\mu^*(A\cap B) +\mu^*(A\cap(Y\cup B^c))$, but I can't justify why I can do that. Any suggestions?? Thank you!
 A: The question asks to prove the following result: 

Let $X$ be a set, $\mu^*$ an outer measure on $X$, and $Y$ a $\mu^*$-measurable subset of $X$ such that 
  $\mu^*(Y)<\infty$. Let $B$ be a subset of  $Y$ such that $\mu^*(Y)=\mu^*(B)+\mu^*(Y\cap B^c)$. 
  Prove that $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ for every $\mu^*$-measurable subset $A$ of $X$.

Proof: Let $A$ be any $\mu^*$-measurable subset of $X$.
In the proof we are going to use the following: 
Since $Y$ is $\mu^*$-measurable, we have 
\begin{align} 
&\mu^*(A) = \mu^*(A \cap Y)+ \mu^*(A \cap Y^c) \:\:\:\: \tag{1} \\&
\mu^*(A\cap B^c) = \mu^*(A\cap B^c \cap Y)+ \mu^*(A\cap B^c \cap Y^c) \:\:\:\: 
\tag{2}
\end{align} 
Since $\mu^*$ is subadditive, we have 
$$\mu^*(Y\cap A^c) \leqslant \mu^*(Y\cap A^c\cap B)+ \mu^*(Y\cap A^c\cap B^c) \:\:\:\: \tag{3}$$
Since $A$ is $\mu^*$-measurable, we have
\begin{align}  
&\mu^*(B)= \mu^*(B\cap A)+ \mu^*(B\cap A^c) \:\:\:\: \tag{4}
\\& \mu^*(Y\cap B^c)= \mu^*(Y\cap B^c\cap A)+ \mu^*(Y\cap B^c\cap A^c) \:\:\:\: \tag{5} 
\\ &\mu^*(Y)= \mu^*(Y\cap A)+ \mu^*(Y\cap A^c) \:\:\:\: \tag{6}
\end{align} 
As one of the hypothesis, we have that
$$\mu^*(Y)= \mu^*(B)+ \mu^*(Y\cap B^c) \:\:\:\: \tag{7}$$
Now, let us prove the result. 
\begin{align}
\mu^*&(A\cap B) +\mu^*(A\cap B^c) =  \\&= \mu^*(A\cap B) + \mu^*(A\cap B^c \cap Y)+ \mu^*(A\cap B^c \cap Y^c) = &\textrm{ by } (2)
\\& = \mu^*(A\cap B) + \mu^*(A \cap Y \cap B^c )+ \mu^*(A\cap Y^c) = &\textrm { used } B\subseteq Y
\\& = \mu^*(A\cap B) + \mu^*(A \cap Y \cap B^c )+ \mu^*(A\cap Y^c) +\\&\:\:\:\:\:+ \mu^*(Y\cap A^c) - \mu^*(Y\cap A^c) \leqslant &\textrm { used } \mu^*(Y)<+\infty
\\& \leqslant  \mu^*(A\cap B) + \mu^*(A \cap Y \cap B^c )+ \mu^*(A\cap Y^c) + \\&\:\:\:\:\: + \mu^*(Y\cap A^c\cap B)+ \mu^*(Y\cap A^c\cap B^c) - \mu^*(Y\cap A^c)=  &\textrm{ by } (3)
\\& =\mu^*(A\cap B) + \mu^*(A \cap Y \cap B^c )+ \mu^*(A\cap Y^c) + \\&\:\:\:\:\:+\mu^*(B\cap A^c)+ \mu^*(Y\cap B^c \cap A^c) - \mu^*(Y\cap A^c)= &\textrm { used } B\subseteq Y
\\& =\mu^*(B) + \mu^*(A \cap Y \cap B^c )+ \mu^*(A\cap Y^c) + \\&\:\:\:\:\:+ \mu^*(Y\cap B^c \cap A^c) - \mu^*(Y\cap A^c)= &\textrm{ by } (4)
\\& =\mu^*(B) + \mu^*(Y \cap B^c )+ \mu^*(A\cap Y^c)  - \mu^*(Y\cap A^c)= &\textrm{ by } (5)
\\& =\mu^*(Y) + \mu^*(A\cap Y^c)  - \mu^*(Y\cap A^c)= &\textrm{ by } (7)
\\& =\mu^*(Y) - \mu^*(Y\cap A^c) + \mu^*(A\cap Y^c)  =
\\& =\mu^*(Y \cap A) + \mu^*(A\cap Y^c)  = &\textrm{ by } (6)
\\& =\mu^*(A)  &\textrm{ by } (1)
\end{align} 
So we have prove that 
$$\mu^*(A\cap B) +\mu^*(A\cap B^c) \leqslant \mu^*(A)$$
Since $\mu*$ is subadditive, we know that 
$$\mu^*(A) \leqslant \mu^*(A\cap B) +\mu^*(A\cap B^c)$$
So we have 
$$\mu^*(A) = \mu^*(A\cap B) +\mu^*(A\cap B^c)$$
Important Remark:  It is worth to note that, if $\mu^*$ is NOT induced by some pre-measure (or measure) $\mu$ (if $\mu^*$ is simply an arbitrary outer measure), the conclusion that $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ for every $\mu^*$-measurable subset $A$ of $X$ is NOT enough to claim that $B$ is $\mu^*$-measurable. In fact, for $B$ to be $\mu^*$-measurable, we should have $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ for every subset ($\mu^*$-measurable or not) $A$ of $X$. 
Example: Let $X=\{a,b,c\}$. Let $\mu^*$ be defined by $\mu^*(\emptyset)=0$, $\mu^*(X)=2$ and $\mu^*(A)=1$, if $A\neq \emptyset$ and $A\neq X$. It is easy to check that $\mu^*$ is an outer measure and that $X$ is $\mu^*$-measurable. Take $Y=X$. We have that $Y$ is $\mu^*$-measurable and $\mu^*(Y)<+\infty$. Let $B=\{a,b\}$. $B$ is a subset of $Y$ and 
$$ \mu^*(Y)= 2 = 1+1 = \mu^*(B)+ \mu^*(\{c\}) = \mu^*(B)+ \mu^*(Y\cap B^c)$$
HOWEVER, taking $A=\{a,c\}$ (which is not $\mu^*$-measurable), we have 
$$ \mu^*(A)= 1 < 1+1 = \mu^*(\{a\})+ \mu^*(\{c\}) = \mu^*(A \cap B)+ \mu^*(A\cap B^c)$$
So, $B$ is not $\mu^*$-measurable.  
Remark 2: In fact, if the outer measure $\mu^*$ is induced by a premeasure defined on a semi-ring $\mathcal H$, then in order to claim that $B$ is $\mu^*$-measurable it is enough to check the condition $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ against all sets $A$ in the $\sigma$-algebra generated by $\mathcal H$, such that $\mu^*(A)<\infty$. 
See: 
Checking Caratheodory-measurability condition on sets of the semiring
In particular, in this case, it is enough to check the condition $\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap B^c)$ against all sets $A$ $\mu^*$-measurable.
