Context: Let $X$ be a set and $\mu$ a measure on a $\sigma-$algebra $\dot{A} \subseteq \mathcal{P}(X)$ and $N$ the $\mu-$Null Set, such that $\forall n \in N,$ $\mu(n)=0$
Define $\bar{A}:= \{A \cup M \in \mathcal{P}(X): A \in \dot{A}, M \in \bar{N}\}$, whereby $\bar{N}:=\{M \in \mathcal{P}(X): \exists n \in N : M \subseteq n\} $
Question: for $B = A \cup M \in \bar{A}$, where $A \in \dot{A}, M \in \bar{N}$, show that $\bar{\mu}(B):=\mu(A)$ is well defined.
Solution: I have great problems understanding the solution that was presented to me. It goes as follows
Let $B = A \cup M = A^* \cup M^*$ where $A, A^* \in \bar{A}$ and $M, M^* \in \bar{N}$. It then goes on to state $\exists n \in N$ such that $M \cup M^*\subseteq n$ (1st Q: What conditions do I have that assure that the UNION $M \cup M^*\subseteq n$?).
Then $\mu(A)=\mu(A \cup n)$ and $\mu(A^*)=\mu(A^* \cup n)$ (2nd Q: Surely this statement would only make sense if $A \cap n = \varnothing$, so disjoint? Otherwise there is no $\sigma-$ additivity, even if $n$ is in a null set???)
Then, it states $A \cup n = A^* \cup n$, and so it follows $\mu(A)=\mu(A^*)$, and therefore it is well-defined. (3rd Q: How do we know $A \cup n = A^* \cup n$?
As stated, I am confused by the solution adopted by the professor. I would be eternally grateful to anyone who could help me understand this solution and answer my aforementioned questions.
We want to show that $\overline{\mu}(B)$ is well-defined. If we define $\overline{\mu}(B) = \mu(A)$, given that $B = A\cup M$, where $A\in\mathcal{P}(A)$ and $M\in\overline{N}$, we see that there may be issues with well-definedness. For example, the interval $[0,1]$ can be written as $(0,1]\cup \{0\}$ or $[0,1)\cup \{1\}$ or $[0,\frac{1}{2})\cup (\frac{1}{2},0]\cup \{0\}$. Perhaps it is possible, that given these different ways of writing $[0,1]$, that $\overline{\mu}([0,1])$ yields a different value, depending on the set $A$. We have to show that $\overline{\mu}$ is invariant of the choice of set.
When we write $B=A\cup M = A^*\cup M^*$, we are testing this invariance. We want to see if, given these different choices of $A$ and $M$, we still get the same measure. To directly address your first question, since $M,M^*\in\overline{N}$,we know that there exists some $n_{M}, n_{M^*}\in N$ such that $M\subseteq n_M$ and $M^*\subseteq n_{M^*}$. Since $n_M$ and $n_{M^*}$ both have null-measure, $n = n_M\cup n_{M^*}$ also has null measure and thus $n\in N$. So $M\cup M^\star\subseteq n\in N$.
Next, $\mu(A) = \mu(A\cup n)$ for any null measure set $n$, regardless of whether or not $A$ and $n$ are disjoint. Using the inclusion-exclusion principle, we have that \begin{align*} \mu(A\cup n) &= \mu(A) + \mu(n) - \mu(A\cap n)\\ &= \mu (A) \end{align*} Another way of seeing this is by appealing to the definition of measure - the infinimum of the sum of the volumes of any rectangular covering - and observing that $n$ can be covered by any set of rectangles with arbitrarily small total volume.
Finally, to see that $A\cup n = A^*\cup n$, we remember how $A, A^*,$ and $n$ were defined. We know $B = A\cup M = A^*\cup M^*$ and thus $A\cup (M\cup M^\star) = A^\star \cup (M\cup M^\star)$. Since $M\cup M^*\subseteq n$, we can see that $A\cup n = A^*\cup n$. This finally leads us to the conclusion that \begin{align*} \mu(A) = \mu(A\cup n) = \mu(A^*\cup n) = \mu(A^*) \end{align*} And thus $\overline\mu(B)$ is well-defined since it doesn't depend on the set decomposition.