Outer measure and Caratheodory's criterion Suppose $m^*$ is an outer measure in Caratheodory's sense on the space $X$, which satisfies $m^*(\emptyset)=0$, $A\subseteq B\implies m^*(A)\le m^*(B)$, and $m^*(\bigcup_n A_n)\le\sum m^*(A_n)$. We define $E$ measurable iff $m^*(E\cap A)+m^*(E^c\cap A)=m^*(A)$ for all $A\subseteq X$, and denote $m(E)=m^*(E)$. Is it true that $m^*$ is always really an outer measure, i.e.
$$m^*(A)=\inf\{\,m(E)\,\vert\,\text{measurable }E\supseteq A\,\}$$
If not, can we put some simple axioms to $m^*$ such that the preceding equality takes place?
I don't know any example of outer measure except Lebesgue outer measure, so I need some help. Thanks.
 A: The only outer measures for which your condition holds are the so-called regular outer measures $^{[1]}$, that is, the ones for which every set $A\subset X$ admits a measurable set $E$ such that $A\subset E$ and $m^\star(A)=m(E)$. (In Munroe's book Introduction to measure and integration, such a set $E$ is called a measurable cover for $A$.)
To prove this assertion we observe that for a regular outer measure the following infimum 
$$\tag{1}\inf \left\{ m(E)\ :\ A\subset E,\ E\text{ is measurable}\right\}$$
is attained at any measurable cover $E$ and $m(E)=m^\star(A)$, by definition. Conversely, if $m^\star(A)$ is given by (1), then$^{[2]}$ we can take a shrinking sequence $E_n$ of measurable sets containing $A$ and such that 
$$\lim_{n\to \infty} m(E_n)=m^\star(A).$$
Then $E=\cap_n E_n$ is a measurable cover for $A$ and $m^\star$ is regular. 
Not every outer measure is regular. As an example, take a set $\Omega$ containing at least two element and define 
\begin{equation}
m^\star(E)=
\begin{cases}
0, & E=\varnothing \\
1, & E\subsetneq \Omega \\
1+\frac{1}{2}, & E=\Omega.
\end{cases}
\end{equation}
This defines an outer measure such that the class of measurable sets is reduced to the trivial sigma algebra $\{\varnothing, \Omega\}$, so any proper subset of $\Omega$ has no measurable cover. 
It is interesting to note that the measure $m$ associated to $m^\star$ is
\begin{equation}
m(E)=\begin{cases} 
0, & E=\varnothing\\
1+\frac{1}{2}, & E=\Omega,
\end{cases}
\end{equation}
and applying the method outlined in the other answer, which is denoted in Munroe's book as a (special case of) Method I, one gets the outer measure 
\begin{equation}
m^\star_0(E)=\begin{cases} 
0, & E=\varnothing \\
1+\frac{1}{2}, & E\subset \Omega.
\end{cases}
\end{equation}
which is different from the original outer measure $m^\star$. 
This phenomenon explains the terminology regular. Regular outer measures are those for which the application of Method I to the induced measure yields the original outer measure. This is the content of Theorem 12.4 at page 99 of Munroe's book.

$^{[1]}$ Not to be confused with the concept of regularity of measures on topological spaces.
$^{[2]}$ As you rightfully point out in your comments above.
A: Given any measure $m$ on a measurable space $(X, \Sigma)$, $m$ may always be extended to an outer measure $m^*$ with the prescription $m^*(E) = \inf \{ m(A): E \subset A, A \in \Sigma\}$.  Note that $m^*(\emptyset) = 0$ is clear since $\emptyset \in \Sigma$.  Further, $m^*$ is clearly monotonic.  The only real think to check is subadditivity.
Given a countable collection $\{A_k\}_k$ of subsets of $X$ and $\epsilon > 0$ (where we'll assume $m^*(A_k) < \infty$ for each k, else there's nothing to prove), we'll pick for each $k$ a set $E_k \in \Sigma$ containing $A_k$ such that $m^*(A_k) + \frac{\epsilon}{2^k} > m(E_k)$.  Then, if $A = \bigcup_k A_k$, we have that $\bigcup_k E_k \in \Sigma$ and contains $A$, so 
$m^*(A) \le m(\bigcup_k E_k) \le \sum_k m(E_k) \le \sum_k m^*(A_k) + \epsilon$ and the result follows.
The last thing to show is that $m^* = m$ on $\Sigma$.  For that, we have for free that $m^*(E) \le m(E)$, for all $E \in \Sigma$ by definition of $m^*$.  Alternatively, if $\epsilon >0$, pick a set $E' \in \Sigma$ containing $E$ with $m^*(E) + \epsilon \ge m(E') \ge m(E)$ (the last inequality following by the monotonicity of $m$), and so we're done.
In this sense, there's really not much of a distinction between measures and outer measures, and many texts in geometric measure theory deal exclusively with outer measures for convenience.
