Alternative characterisations of the Outer Measure $ \mu^*$ In my lectures on Measure Theory, I came across the notion of an outer measure. The way in which this was presented to me was quite confusing and understanding this was quite difficult.
Having done some extra reading on this, I have found that there is a much simpler way to characterise outer measures that makes a lot of sense to me.
The definition that makes sense to me is the following:

An Outer Measure is a function $\mu^* : P(\Omega) \rightarrow [0, + \infty]$ such that

*

*$\mu^* (\emptyset) = 0$

*$A \subseteq B \implies \mu^* (A) \leq \mu^* (B)$

*$A_1, A_2, A_3, ... \in P(\Omega) \implies \mu^* (\cup _{n=1}^{\infty}A_n) \leq \sum _{n=1}^{\infty} \mu^* (A_n)$

After reading this definition and some surrounding explanation, this made a lot of sense to me.
However, the definition presented in my lectures is quite different and I am confused about whether or not these represent the same thing (and why they have decided to characterise the Outer Measure in this way). They use the following definition:

An Outer Measure is a function $\mu^* : P(\Omega) \rightarrow [0, + \infty]$ such that for a countably additive measure $\mu$ and the set $D_E := \{(A_n)_{n \in \mathbb{N}} : A_i \in \mathscr{A}$ and $E \subseteq \cup_{i=1}^{\infty} A_i \}$ we defined
$\mu^*(E) := \inf\limits_{(A_n) \in D_E} \sum _{i=1}^{\infty} \mu(A_n)$

This definition is very unclear to me, however, it is the only definition that has been presented to me in my lectures and so I don't want to try to build up an intuition for the previous definition if these are different in some way.
I would be grateful for some clarification here.
 A: Suppose that $\Omega$ has at least three points and consider the function $\mu^\ast \colon P(\Omega) \to [0, \infty]$ defined by
$$
\mu(A) = \begin{cases}
0 & \text{if }A = \emptyset \\
1 & \text{if } A \neq \emptyset \text{ and } A \neq \Omega \\
3 & \text{if } A = \Omega
\end{cases}
$$
It is not difficult to prove that $\mu^\ast$ is an outer measure (in the first sense). Suppose by contradiction that there is a $\sigma$-algebra $\mathscr{A}$ and a measure $\mu \colon \mathscr{A} \to [0, \infty]$ such that $\mu^\ast$ derives from $\mu$ as in the second definition. Then we assert that $\mu^\ast(A) = \mu(A)$ for all $A \in \mathscr{A}$ (this is an easy consequence of the second definition). It implies that $\mathscr{A} = \{\emptyset, \Omega\}$. Indeed, if there is $A \in \mathscr{A} \setminus \{\emptyset, \Omega\}$, then
$$3 = \mu^\ast(\Omega) = \mu(\Omega) = \mu(A) + \mu(\Omega \setminus A) = \mu^\ast(A) + \mu^\ast(\Omega \setminus A) = 2$$
Next, the formula which gives $\mu^\ast$ in terms of $\mu$ implies that $\mu^\ast(A) = 3$ for all nonempty $A \subset \Omega$.
Therefore, the second definition is more restricted. In fact, we say that an outer measure $\mu^*$ is regular if for all $A \subset \Omega$, there exists a $\mu$-measurable set $B \supset A$ (in the sense of Carathéodory) such that $\mu^*(A) = \mu^*(B)$. An outer measure $\mu^*$ satisfies the second definition if and only if it is regular, see for example Federer's Geometric measure theory.
