Let $X$ be a set, $S=\{\emptyset,X\}$, and define $\mu:S\to[0,\infty)$ by $\mu(\emptyset)=0$ and $\mu(X)=1$. Determine the outer measure $\mu^*$ induced by $\mu$, and find the $\sigma$-algebra of measurable sets.
I know that outer measure $\mu^*$ must satisfy the properties that (1) $\mu^*(\emptyset)=0$, and (2) that $\mu^*$ is countably monotone, but I am unsure of how to construct $\mu^*$ from $\mu$.
Royden defines the outer measure induced by a set function $\mu : S \to [0,\infty]$ (where $S$ is an arbitrary collection of subsets of $X$) as follows: set $\mu^*(\emptyset) = 0$, and for any nonempty subset $E \subseteq X$, put $$\mu^*(E) = \inf\left\{\sum_{k=1}^{\infty}\mu(E_k) : E \subseteq \bigcup_{k=1}^{\infty}E_k,\ \ E_k \in S\right\}$$
Let's give a name to the set in braces; I'll call it $A$. Note that $A$ is a subset of $[0, \infty]$.
In this problem, we have $S = \{\emptyset, X\}$, so in the union $\bigcup_{k=1}^{\infty}E_k$, each $E_k$ is either $\emptyset$ or $X$.
Let $E$ be any nonempty subset of $X$. (I'm assuming that $X$ itself is nonempty.)
Note that $\sum_{k=1}^{\infty}\mu(E_k)$ is either an integer or $\infty$, since the summands are all either $0$ or $1$. The only way the sum can be zero is if $E_k = \emptyset$ for all $k$, in which case $\bigcup_{k=1}^{\infty}E_k = \emptyset$, which does not contain $E$ as a subset.
So, $A$ doesn't contain $0$, hence its infimum must be at least $1$.
In fact, $A$ contains $1$, so its infimum is exactly $1$. To see this, let $E_1 = X$ and $E_k = \emptyset$ for all $k > 1$. Then $\bigcup_{k=1}^{\infty}E_k = X \cup \emptyset \cup \emptyset \cup \cdots = X$, which certainly contains $E$ as a subset. Therefore $A$ contains $\sum_{k=1}^{\infty}\mu(E_k) = \mu(X) + \mu(\emptyset) + \mu(\emptyset) + \cdots = 1 + 0 + 0 + \cdots = 1$.
We conclude that $\mu^*(E) = 1$ for any nonempty subset $E \subseteq X$, and of course $\mu^*(\emptyset) = 0$.
For the next step we want to know which $E \subseteq X$ satisfy the Caratheodory criterion $$\mu^*(B) = \mu^*(B \cap E) + \mu^*(B \cap E^c)$$ for all $B \subseteq X$.
In particular, setting $B = X$, we see that such a set $E$ must satisfy $$1 = \mu^*(X) = \mu^*(E) + \mu^*(E^c)$$ The terms on the RHS are either $0$ or $1$, so exactly one of them must be $0$ and the other must be $1$. In particular, since one of the terms must be zero, this means that either $E$ or $E^c$ must be $\emptyset$, and consequently either $E^c$ or $E$ must be $X$.
The above reasoning shows that the only subsets of $X$ which satisfy the Caratheodory criterion with respect to $\mu^*$ are $\emptyset$ and $X$, i.e. $S = \{\emptyset, X\}$ is the $\sigma$-algebra of measurable sets with respect to the measure induced by $\mu^*$.