Extending a measure I need help with the following:
Let $(X, \mathcal{A}, \mu)$ be finite measure space and $D \subseteq X$ such that $D \notin \mathcal{A}$. 
a) Check that $\sigma(\mathcal{A} \cup \{D\})=\mathcal{A}_D:=\{(A\cap D) \cup (B \cap D^C): A, B \in \mathcal{A}\}.$
b) Show for a set $E \in \mathcal{A}_D$ with representation $E=(A \cap D) \cup (B \cap D^C)$ that $\mu_D(E):=\mu(A \cap M) + \mu(B \cap M^C)$ is well-defined (although $A,B,M$ are not necessarily unique), where $M \in \mathcal{A}$ is a measurable hull of D, so $D \subseteq M$ and $\mu^*(D)=\mu^*(M)$ and $\mu^*$ is the outer measure generated by $\mu$.
c) Show that $\mu_D: \mathcal{A}_D \rightarrow [0, \infty]$ is a measure with $\mu_D(D)=\mu^*(D)$ which extends $\mu$.
For (a), I thought that "$\supseteq$" is clear and for "$\subseteq$" I shouwed for exapmle:
$$\bigcup_i ((A_i \cap D) \cup (A_i' \cap D^C))=((\bigcup_i A_i \cap D) \cup (\bigcup_i A_i') \cap D^C))$$
where the right hand side is contained in $\mathcal{A}_D$. The same can be shown for the intersection or for taking complements.
For (b), we had the proposition, that $\forall D \subseteq X \text{ } \exists M \in \mathcal{A}: D \subseteq M \text{ and } \mu^*(D)=\mu^*(M)$. Can I use this to show the claim of b?
Can someone help me or give me a hint?
 A: Your solution for (a) seems good. The basic idea is to show that: every element of $\mathcal{A}_D$ belongs to $\sigma(\mathcal{A}\cup\left\{D\right\})$ (which gives the inclusion $\supseteq$), and that $\mathcal{A}_D$ is indeed a $\sigma$-algebra containind $\mathcal{A}$ and $D$ (which gives the inclusion $\subseteq$).
For (b), you note that $\mu_D(E)$ depends on $A$, $B$ and $M$. In the very definition of $\mu_D$, you are already using the result of existence of a measurable hull, so you shouldn't really worry about that.
Now, to solve (b), I think the easier way is to first prove a lemma, which is basically the case where $M=X$.

Lemma: Let $(Y,\mathcal{M},\mu)$ be a finite measure space and $E\subseteq X$ be a subset with $\mu^*(E)=\mu(X)$. Consider the $\sigma$-algebra $\mathcal{M}_E=\left\{A\cap E:A\in\mathcal{M}\right\}$. Then $\mu^*(A\cap E)=\mu(A)$ for all $A\in\mathcal{A}$, hence the mapping $\mu_E(A\cap E)=\mu(A)$ ($A\in\mathcal{M}$) is well-defined, and it is a measure on $\mathcal{M}_E$.
Proof: Let $A\in\mathcal{M}$. Given any covering $\left\{C_n\right\}_{n=1}^\infty$, $C_n\in\mathcal{M}$, of $A\cap E$ (i.e., $A\cap E\subseteq \bigcup_n C_n$), then $\left\{C_n\right\}\cup\left\{Y\setminus A\right\}$ covers $E$, hence, by definition of the outer measure $\mu^*$,
$$\mu(Y)=\mu^*(E)\leq\mu(Y\setminus A)\sum_n \mu(C_n)=\mu(Y)-\mu(A)+\sum_n\mu(C_n),$$
so $\mu(A)\leq\sum_n\mu(C_n)$. Taking the infimum on the coverings $\left\{C_n\right\}$, we obtain $\mu(A)\leq \mu^*(A\cap E)$, and the reverse inequality is clear.
Hence, the mapping $\mu_E(A\cap E)=\mu^*(A\cap E)=\mu(A)$ is well-defined, clearly $\mu_E(\varnothing)=0$ and, if $\left\{A_n\right\}$ is a sequence of elements of $\mathcal{M}$ with the $A_n\cap E$ disjoint, then
\begin{align*}
\mu_E\left(\bigcup_{n=1}^\infty A_n\cap E\right)&=\mu\left(\bigcup_{n=1}^\infty A_n\right)=\mu\left(\bigcup_{n=1}^\infty (A_n\setminus \bigcup_{j=1}^{n-1} A_j)\right)\\
&=\sum_{n=1}^\infty \mu\left(A_n\setminus\bigcup_{j=1}^{n-1}A_j\right)=\sum_{n=1}^\infty\mu_E\left((A_n\setminus\bigcup_{j=1}^{n-1}A_j)\cap E\right)\\
&=\sum_{n=1}^\infty\mu_E\left((A_n\cap E)\setminus\bigcup_{j=1}^{n-1}(A_j\cap E)\right)=\sum_{n=1}^\infty \mu_E(A_n\cap E),
\end{align*}
so $\mu_E$ is a measure. Q.E.D.

(b) Here we should take a little care: Let $\mathcal{A}_M=\left\{A\in\mathcal{A}:A\subseteq M\right\}$. Note that $\mu$ restricted to $\mathcal{A}_M$ is a measure, which I'll denote $\mu_M$. The outer measure (on $M$) induced by $\mu_M$ is simply the outer measure $\mu^*$ restricted to subsets of $M$. If this is not clear, you should check this. This fact allows us to use the lemma above.
Let's show that $\mu_D$ as in the question independs on $A$ and $B$ (but at first we consider some measurable hull $M$ fixed). Suppose that $A,B,A',B'$ are sets in $\mathcal{A}$ with
$$(A\cap D)\cup(B\setminus D)=(A'\cap D)\cup(B'\setminus D).$$
If we take intersection with $M^c$ in the equality above, we obtain $B\setminus M=B'\setminus M$. On the other hand, if we intersect the equality above with $D$, we obtain $A\cap D=A'\cap D$. From the lemma, (with $D$ in place of $E$ and $M$ in place of $Y$),
$$\mu(A\cap M)=\mu^*(A\cap M\cap D)=\mu^*(A\cap D)=\mu^*(A'\cap D)=\cdots=\mu(A'\cap D)$$
therefore $\mu(A\cap M)+\mu(B\setminus M)=\mu(A'\cap M)+\mu(B'\setminus M)$, so $\mu_D$ independs of $A$ and $B$.
Now, suppose we had another measurable hull $M'$ of $D$. Then $M\cap M'$ covers $D$, hence
$$\mu(M)=\mu^*(D)\leq\mu(M\cap M')=\mu\left(M\setminus(M\setminus M')\right)=\mu(M)-\mu(M\setminus M')$$
so $\mu(M\setminus M')=0$. For all $A\in\mathcal{A}$,
$$\mu(A\cap M)=\mu(A\cap M\cap M')+\mu((A\cap M)\setminus M')\leq\mu(A\cap M')+\mu(M\setminus M')=\mu(A\cap M'),$$
and simetrically $\mu(A\cap M')\leq\mu(A\cap M)$. Similarly you show that $\mu(B\setminus M)=\mu(B\setminus M')$ for all $B\in\mathcal{A}$. This shows that $\mu_E$ also independs of $M$.
I'm just going to give some hints on part (c): If $\left\{(A_i\cap D)\cup(B_i\setminus D)\right\}$ is a disjoint family in $\mathcal{A}_D$, then the $(A_i\cap D)$ are all disjoint, and the $(B_i\setminus D)$ are also disjoint. You should use the equality you proved on item (a) to write the union as an element of $\mathcal{A}_D$. Using the lemma, you can obtain $\mu(\bigcup A_i\cap M)=\sum_i\mu(A_i\cap M)$. The analysis of the $B_i$'s is easier. Putting all together, you'll conclude that $\mu_D$ is a measure.
