Let $\mu$ be a finite measure on $(X,M)$, and let $\mu^*$ be the outer measure induced by $\mu$. Suppose that $E\subset X$ satisfies $\mu^*(E) = \mu^*(X)$ (but not that $E\in M$).
a.) If $A,B\in M$ and $A\cap E = B\cap E$ then $\mu(A) = \mu(B)$
b.) Let $M_{E} = \{ A \cap E: A\in M\}$, and define the function $\nu$ on $M_E$ defined by $\nu(A\cap E) = \mu(A)$ (which makes sense by (a)). Then $M_E$ is a $\sigma$-algebra on $E$ and $\nu$ is a measure on $M_E$.
Attempted proof a.) Suppose $A,B\in M$ and $A\cap E = B\cap E$ where $E\subset X$. Since $\mu$ is finite on $(X,M)$ then $\mu(X) < \infty$ and we have $\mu(X) = \mu(A\cup A^c) = \mu(B\cup B^c)$. Note that $$\mu(A) + \mu(A^c\cap B) = \mu(A\cup B) = \mu(B) + \mu(B^c\cap A)$$ by symmetry, so it suffices to show that $\mu(A^c\cap B) = 0$.
We have that $E\subset A\cup B^c = (A^c\cap B)^c$. So $$\mu(X) = \mu^*(X) = \mu^*(E)\leq \mu^*((A^c\cap B)^c) = \mu((A^c\cap B)^c) = \mu(X) – \mu(A^c\cap B)$$ and hence $\mu(A^c\cap B) = 0$.
Attempted solution b.) Let $M_E = \{A\cap E: A\in M\}$ and define a function $\nu$ on $M_E$ defined by $\nu(A\cap E) = \mu(A)$. To show that $M_E$ is a $\sigma$-algebra I believe we have to first show that it is an algebra and in doing so we have to show that $M_E$ is closed under finite unions and complements. Then once we have shown that $M_E$ is an algebra we have to show that it is closed under countable unions or countable intersections to show that it is a $\sigma$-algebra.
I am stuck here any suggestions on this part is greatly appreciated.
Best Answer
Before we proceed to the proof, it is worth to understand the idea in this exercise. Given a finite measure $\mu$ on $(X,M)$, and $E \subseteq X$, we want to "restrict $\mu$ to $E$".
Suppose $E\in M$. First we need to restrict $M$ to $E$. The way to do it is very natural, we define $$M_{E} = \{ A : A\in M \text{ and } A\subseteq E\}$$ and we define, for all $A\in M_E$, $\nu(A)=\mu(A)$. It is easy to prove that $M_E$ is a a $\sigma$-algebra and $\nu$ is a measure.
Now, what happens if $E$ is not measurable ($E\notin M$)? In this case, $M_E$ as previously defined may reduce to just $\{\emptyset\}$ and $E$ surely will not be in $M_E$ ($M_E$ is no longer a $\sigma$-algebra). So, "restriction of a measure" to a non-measurable set, in general, does not work.
However, IF $\mu^*(E) = \mu^*(X)$, we can adjust our definitions and have a similar result. We define $$M_{E} = \{ A \cap E: A\in M\}$$ and, for all $A\cap E\in M_E$, $\nu(A\cap E)=\mu(A)$ Then we need to prove that $M_{E}$ is a $\sigma$-algebra and that $\nu$ is a measure. The first step to prove $\nu$ is a measure is to prove that $\nu$ is well defined. So we must prove that, if $A, B\in M$ and $A \cap E =B \cap E$, then $\nu(A \cap E)=\nu(A \cap E)$, that is, $\mu(A)=\mu(B)$. (That is why we need item a. in the exercise).
Now let us proceed to the proof.
Proof:
a.) Suppose $A,B\in M$ and $A\cap E = B\cap E$. Then $$(A \setminus B) \cap E = A\cap B^c \cap E = (A\cap E) \cap B^c = (B\cap E) \cap B^c = \emptyset$$ So $A \setminus B \subseteq E^c$, so $E \subseteq (A \setminus B )^c$, and we have, using that $(A \setminus B )^c\in M$ ,
$$\mu(X)=\mu^*(E)\leqslant \mu^*((A \setminus B )^c) = \mu((A \setminus B )^c)\leqslant \mu(X)$$ So, we have $ \mu((A \setminus B )^c) =\mu(X)$.
On the other hand, we have that $\mu(X)=\mu(A \setminus B ) +\mu((A \setminus B )^c)$. So we get $$\mu(X)=\mu(A \setminus B ) +\mu(X)$$ Since $\mu(X)<\infty$, we have $ \mu(A \setminus B )=0$
In a similar way, we can prove that $ \mu(B \setminus A )=0$.
Now, note that $$\mu(A)=\mu(A)+0=\mu(A)+\mu(B \setminus A )=\mu(A\cup B ) = \mu(B)+\mu(A \setminus B )= \mu(B)+0=\mu(B)$$ So, we have $\mu(A)=\mu(B)$.
b.) It is straight forward to check that $M_E$ is a $\sigma$-algebra on $E$. From item a, we know that $\nu$ is well defined. It is straight forward to check it is in fact a measure.
$M_{E} = \{ A \cap E: A\in M\}$ is a $\sigma$-algebra on $E$.
i. Since $\emptyset \in M$, $\emptyset = \emptyset \cap E \in M_{E}$.
ii. Given any $B\in M_{E}$, there is $A \in M$ such that $B=A \cap E$ then $$E\setminus B = E \setminus (A\cap E) = (X\setminus A) \cap E$$ Since $X\setminus A \in M$, we have that $E\setminus B \in M_{E}$.
iii. Given any countable family $\{B_n\}_{n\in \mathbb{N}}$ of sets in $M_{E}$, for each $n\in \mathbb{N}$, there is $A_n \in M$ such that $B_n=A_n \cap E$. So $$ \bigcup_{n\in \mathbb{N}} B_n= \bigcup_{n\in \mathbb{N}} (A_n \cap E) = \left( \bigcup_{n\in \mathbb{N}} A_n \right)\cap E$$ Since $\bigcup_{n\in \mathbb{N}} A_n \in M$, we have that $\bigcup_{n\in \mathbb{N}} B_n \in M_{E}$.
So we have proved $M_{E}$ is a $\sigma$-algebra on $E$.
Now, let us prove that the function $\nu$ on $M_E$ defined by $\nu(A\cap E) = \mu(A)$ is a measure on $M_E$.
First, note that, as a consequence of item a, $\nu$ is well defined as a function.
Since $\emptyset=\emptyset \cap E$, we have $$\nu(\emptyset)=\nu(\emptyset \cap E)=\mu(\emptyset)=0$$
Second, given any countable family $\{B_n\}_{n\in \mathbb{N}}$ of disjoint sets in $M_{E}$, for each $n\in \mathbb{N}$, there is $A_n \in M$ such that $B_n=A_n \cap E$. Note that the sets $A_n$ may not be disjoint.
Let, for each $n\in \mathbb{N}$, define $C_{n}=A_{n} \setminus \bigcup_{i=0}^{n-1} A_i$, (note that $C_0=A_0$). Then, we have that the sets $C_n$ are disjoint sets in $M$, and we also have
$$ C_n\cap E= \left( A_{n} \setminus \bigcup_{i=0}^{n-1} A_i \right)\cap E= (A_{n}\cap E) \setminus \bigcup_{i=0}^{n-1} (A_i \cap E)= B_n \setminus \bigcup_{i=0}^{n-1} B_i = B_n$$ the last step holds because $\{B_n\}_{n\in \mathbb{N}}$ is a family of disjoint sets.
So, by the definition of $\nu$, for all $n\in \mathbb{N}$, $$\nu(B_n) = \nu(C_n\cap E)= \mu(C_n) \tag{1}$$ and we also have that $\bigcup_{n\in \mathbb{N}} C_n \in M$ and $$\bigcup_{n\in \mathbb{N}} B_n = \bigcup_{n\in \mathbb{N}} (C_n \cap E)=\left ( \bigcup_{n\in \mathbb{N}} C_n \right) \cap E$$ So we have, using $(1)$, $$\nu \left ( \bigcup_{n\in \mathbb{N}} B_n \right) = \nu \left (\left ( \bigcup_{n\in \mathbb{N}} C_n \right) \cap E\right)= \mu\left ( \bigcup_{n\in \mathbb{N}} C_n \right) = \sum_{n\in \mathbb{N}} \mu(C_n)=\sum_{n\in \mathbb{N}} \nu(B_n) $$