TITLE: Understanding when a set is outer-measurable
QUESTION [2 upvotes]: Suppose $\mu$ is an outer measure on some arbitrary set $X$. We say a subset $E$ of $X$ is $\mu$-measurable if and only if
$$\mu(A) = \mu(A \cap E) + \mu(A \setminus E), \quad \forall A\subset X.$$
I am having some trouble understanding the motivation behind this statement. Wikipedia gives one interpretation, which is to think of a $\mu$-measurable subset as a building block "breaking any other subset apart into pieces (namely, the piece which is inside of the measurable set together with the piece which is outside of the measurable set)". Is the motivation behind the statement simply to rule out ill-behaved sets such as the Vitali set and nothing else?
I suppose some of the trouble I am having is, despite knowing about examples such as the Vitali set, I still am having trouble visualizing a set which does not satisfy this condition. How does this condition help obtain a proper measure from an outer measure?

REPLY [5 votes]: Since I have the copyright on an explanation that might help I can repost it here.
The problem that most students encounter for this one is that they have aleady see the Lebesgue definition using outer and inner measures.  But that approach does not work in general situations.  The Greek mathematician Constantin Carathéodory (1873–1950) was the one who developed the correct approach.  But why does it work and why doesn't the simpler inner/outer measure approach work?
This example from the reference [1] was meant to illustrate.
Example 2.29 Let $X = \{1,2,3\}$. Let $μ^∗( \emptyset) = 0$, $μ^∗(X) = 2$, and $μ^∗(A) = 1$ for every
other set $A⊂X$. It is a routine matter to verify that $μ^∗$ is an outer measure.
Suppose now
that we wish to mimic the procedure that worked so well for the Peano–Jordan content and the Lebesgue measure. We could take our cue from the formula in assertion 2.4 and define a version of the inner measure for this example as
$$μ_∗(A) = μ^∗(X)−μ^∗(X \setminus A) = 2−μ^∗(X \setminus A).$$
If we then call $A$ measurable provided that $μ^∗(A) = μ_∗(A)$, and let
$μ(A) = μ^∗(A)$.
for such sets, our process is complete. We find that all eight subsets of X are measurable
by this definition, but $μ$ is clearly not additive on $2^X$ . The classical inner–outer measure procedure completely fails to work in this simple example!
A bit of reflection pinpoints the problem. The inner–outer measure approach puts a
set $A$  to the following test stated solely in terms of $μ^∗$:
Is it true that
$μ^∗(A)+μ^∗(X \setminus A) = μ^∗(X)$?
[Here]  every $A ⊂ X$ passed this test. But, for $A = \{1\}$ and $E = \{1,2\}$, we see
that
$$μ^∗(A)+μ^∗(E \setminus A) = 2 > 1 = μ^∗(E).$$
Thus, while $μ^∗$ is additive with respect to $A$ and its complement in $X$, it is not with
respect to $A$ and its complement in $E$.

REFERENCE:
[1] http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf
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