I'm reading Terrence Tao's An Introduction to Measure Theory. On pp. 20, he writes,
In analogy with the Jordan theory, we would also like to define a concept of “Lebesgue inner measure” to complement that of outer measure. Here, there is an asymmetry (which ultimately arises from the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner measure by replacing finite unions of boxes with countable ones.
I don't quite get this argument. The definition of the Jordan Inner Measure is:
For a bounded set, $A$, in $\mathbb{R}^d$,
\begin{align} m_{* (J)}(A) = \sup_{E \subset A} m(A), \end{align} where $E$ is an elementary set in $\mathbb{R}^d$ definite to be the finite union of boxes, which are just $d-$dimensional Cartesian products of intervals in $\mathbb{R}$.
If we replace finite unions by countable unions, we can define, \begin{align} m_{* (L)}(A) = \sup_{E \subset A} m(A), \end{align} where $m_{*(L)}(A)$ is the Lebesgue inner measure.
Clearly, we have,
\begin{align} m_{*(J)}(A) \leq m_{*(L)}(A). \end{align}
This is analogous to the result,
\begin{align} m^{*(L)}(A) \leq m^{*(J)}(A). \end{align}
Since it's not too hard to show that,
\begin{align} m_{*(L)}(A) \leq m^{*(L)}(A), \end{align}
we have,
\begin{align} m_{*(J)}(A) \leq m_{*(L)}(A) \leq m^{*(L)}(A) \leq m^{*(L)}(A). \end{align}
Hence the Lebesgue measure can be though of as a refinement of the Jordan measure.
In light of this discussion, why does Tao dismiss the idea of defining the Lebesgue measurable sets as those for which the inner and outer measures coincide?
Here are the link to the notes:
https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf