$m_*$ is not subadditivity

It is easy to prove that Lebesgue outer measure is subadditivity, namely, for any set $E_1,E_2,\dots$ we have \[
m^*\bigg(\cupp_{n=1}^\infty E_n\bigg) \leq \sum_{n=1}^\infty m^*(E_i),
\] therefore so is Lebesgue measure since \[
m=m^*|_{\{\text{measurable set}\}}.
\] One may ask whether Lebesgue inner measure is also subadditive, the answer is NO. This explains why there is no such result in any textbook on real analysis/measure theory.

To mention this, let's go through the following standard exercise:

Exercise. Suppose $A_1,A_2,\dots$ are pairwise disjoint subsets of $\R$, show that \[
m_*\bigg(\cupp_{i=1}^\infty A_i\bigg)\ge \sum_{i=1}^\infty m_*(A_i).
\]
Therefore if $m_*$ were subadditive, then we would get a countably additive set function on $2^\R$ that extends the standard length function on interval, but this is impossible by Ulam's Theorem:

This result is extracted from An Introduction to Measure And Integration by Inder K. Rana. Knowledge in axiomatic set theory will be needed in this proof.