Im in a trouble. I want to prove this proposition:
Let $A,B \subseteq \Re^n $ such that $d(A,B)>0$. Then, $ m_*(A\cup B) \le m_*(A) + m_*(B) $ . Where $m_*$ is the Lebesgue inner measure.
I don't know if this lemma is true. But it would help me if it were. Thank you! $<2+1$
Your statement is true.
Recall that $$m_{*}(A) = \sup_{U\subset A, U \textrm{meas}} m(U)= \sup_{K\subset A, K \textrm{compact}}m(K) $$
Let $d = d(A,B)$. Consider $A_1 = \{x \ | d(x,A) \le \frac{d}{3}\}$, $B_1= \{x \ | d(x,B) \le \frac{d}{3}\}$. We have $A_1$, $B_1$ closed and disjoint, $A\subset A_1$, $B\subset B_1$. This measurable separation of $A$, $B$ is all we need.
Take $U\subset A\cup B$, $U$ measurable. We have $U\cap A_1 \subset A$, $U\cap B_1 \subset B$ and $U = (U\cap A_1) \cup U\cap B_1 $ a partition of $U$.
We have $$m(U) = m(U\cap A_1) + m(U \cap B_1) \le m_{*}(A) + m_{*}(B)$$
Now take the supremum over all $U$ measurable, $U\subset A\cup B$.
Note that in general for $A$, $B$ disjoint we have $$m_{*}(A)+m_{*}(B) \le m_*(A\cup B)$$ Therefore, if $A$, $B$ are moreover separated, we have equality.