Let $X$ be any set containing more than one point and $A$ a proper
nonempty subset of $X$. Define $S= \{A, X\}$ and the set function
$\mu: S \rightarrow [0, \infty]$ by $\mu(A) = 1$ and $\mu(X) = 2$.
Show that $\mu: S \rightarrow [0, \infty]$ is a premeasure. Can $\mu$
be extended to a measure? What are the subsets of $X$ that are
measurable with respect to the outer measure $\mu^*$ induced by $\mu$?How would these change if instead we considered the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $R$ and define $\mu(\emptyset)=0$, $\mu([0, 1])=1$, $\mu([0, 3])=1$, $\mu([2, 3])=1$?
Solution 1:
Take $a \in A$, then $\mu^*(\{a\}) = \inf \sum_{k=1}^\infty \mu(E_k)$, $\{a\} \subset\bigcup_{k=1}^\infty E_k$, where $E_k \in S$, then $\mu^*(\{a\}) = 1$, and, similarly, $\mu^*(\{b\}) = 2$ where $b \in X$.
Let $\emptyset \neq C \subset A$. Then, $\mu^*(C) = 1$. Suppose, $D \cap (X\sim A) \neq \emptyset$, then $\mu^*(D)=2$.
Show that $\mu: S \rightarrow [0, \infty]$ is a premeasure.
Using the definition of premeasure:
Let $S$ be a collection of subsets of a set $X$ and $\mu: S \rightarrow [0, \infty]$ a set function. Then, $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone, and if $\emptyset$ belongs to $S$, then $\mu(\emptyset)=0$.
Should I proceed the same way as in Lebesgue Measure?
Finitely Additive: So, show $\mu(\bigcup_{k=1}^n E_k) \leq \sum_{k=1}^n \mu (E_k)$ (by subadditivity) and then, $\mu(\bigcup_{k=1}^n E_k) \geq \sum_{k=1}^n \mu (E_k)$.
Countably Monotone: $\mu(E) \leq \sum_{k=1}^\infty \mu(E_k)$
or should I do something else in general measure?
Lastly: Since $\emptyset \notin S$, this isn't necessary.
Can $\mu$ be extended to a measure?
Is $A$ measurable?
If so, then $\mu^*(B) = \mu^*(B \cap A) + \mu^*(B \cap A^c)$.
Let $B = \{a, b\}$. $\mu^*(B) = 2$ (from before).
$\{a\} \in B \cap A \subset A$, $\mu^*(A \cap B) = 1$.
$\{b\} \in B \cap A^c$, $\mu^*(B \cap A^c) = 2$
$2 \neq 1 + 2$
Is this enough to show that A is not measurable and that $\mu$ cannot be extended to measure?
Which subsets of $X$ are measurable with respect to the outer measure $\mu^*$ induced by $\mu$?
So, we need to find the $\mu^*$ measurable sets.
$E$ is measurable with respect to $\mu^*$ if and only if $\forall B \subset X$.
$$\mu^*(B) = \mu^*(B \cap E) + \mu^*(B \cap E^c)$$
Then, $\emptyset, X$ are measurable because
$$\mu^*(B \cap\emptyset) + \mu^*(B \cap S) = \mu^*(\emptyset) + \mu^*(B) = \mu^*(B)$$
and
$$\mu^*(B \cap X) + \mu^*(B \cap (S \sim B)) = \mu^*(B) + \mu^*(\emptyset) = \mu^*(B)$$
Solution 2: I won't go into anything because it seems like it will basically be the same thing. Are there any major differences that I should know about?
Best Answer
You have a strange definition of premeasure. Your system of sets is not a ring, not even a semi-ring. Therefore there is not much to prove: Finite additivity is vacuous, because there are no disjoint sets in the system, and similarly for countable subadditivity.
What does it mean for $\mu$ to be extendable to a measure? Does it mean that there exists a measure on the $\sigma$-algebra generated from $S$ which equals $\mu$ when restricted to $S$? If so, then yes: We have $\sigma(S) = \{\emptyset, A, X\setminus A, X\}$ and can set $\mu(\emptyset) = 0$ and $\mu(X \setminus A) = 1$.
$\emptyset$ and the whole base set are always measurable with respect to the outer measure. But the question is whether there exist more measurable sets. You have already proved that $A$ is not measurable. What about other subsets of $X$?
For 2.: No, the steps are the same.