Let $A,B\subset\mathbb R$. Then $m^*(A\cap B)+m^*(A\cup B)\le m^*(A)+m^*(B)$
It holds that $m^*(A\cup B)\le m^*(A)+m^*(B)$ for any $A,B$
hence $A\cap B$ must be the empty set.
how to prove that $(A\cap B)$ is the empty set?
I thought in using complements but I arrive to $m^*((A^c\cup B^c)^c)$
though I don't see how would that help.
I also know think both sets are Lebesgue measure
Someone please help me
It's not true in general that $m^*(A \cap B)=0$. Consider $A=[0,1]$ and $B=[-1,1]$.
Edit: Since $A$ and $B$ are Lebesgue measurable, we have $$m^*(A) = m^*(A\setminus B) + m^*(A \cap B)$$ and $$m^*(B) = m^*(B \setminus A) + m^*(A\cap B).$$ So, $$m^*(A) + m^*(B) = m^*(A\setminus B) + m^*(A \cap B) + m^*(B \setminus A) + m^*(A\cap B) = m^*(A\cup B) + m^*(A \cap B).$$ The final equality holds as $A\setminus B \cup B\setminus A \cup (A\cap B) = A \cup B$.
Edit 2: I took the OP to indicate that $A$ and $B$ are Lebesgue measurable. This proof WON'T work without that assumption.
Edit 3: Now we need to try to extend this to arbitrary $A,B \in \mathcal{P}(\mathbb{R})$. To do so, we need a definition of Lebesgue outer measure: $$m^*(E) = \inf\bigg\{ \sum_i \ell(I_i) : I_i \text{ open intervals and } \bigcup_i I_i \supset E \bigg\},$$ where $\ell$ denotes length. Let $\varepsilon > 0$ be arbitrary. Then, by definition, there exists $I = \cup_i I_i$ and $J = \cup_j J_j$, unions of open intervals with $I \supset A$ and $J \supset B$, so that $m^*(A)\geq m^*(I) - \varepsilon$ and $m^*(B) \geq m^*(J) - \varepsilon$. Then, noting that $I$ and $J$ are (Lebesgue) measurable, we have $$m^*(A) + m^*(B) \geq m^*(I) + m^*(J) - 2\varepsilon = m^*(I\cap J) + m^*(I \cup J) - 2\varepsilon \geq m^*(A \cap B) + m^*(A \cup B) - 2\varepsilon.$$ Sending $\varepsilon \downarrow 0$ will complete the proof.
Edit 4: Based on a comment, I'm going to offer a bit of extra detail as to why $m^*(A\cap B) + m^*(A \cup B) = m^*(A) + m^*(B)$ when $A$ and $B$ are (Lebesgue) measurable sets. Utilizing the Caratheodory condition on $A$ and $B$ we arrived at $$m^*(A) + m^*(B) = m^*(A\setminus B) + m^*(A \cap B) + m^*(B \setminus A) + m^*(A\cap B).$$ We want to get $m^*(A \cap B) + m^*(A \cup B)$ out of this expression, so let's look at the terms $m^*(A\setminus B) + m^*(B \setminus A) + m^*(A\cap B)$. Consider the following union: $$A\setminus B \cup B\setminus A \cup A\cap B = A \cup B.$$ Notice that the three sets on the LHS are all mutually disjoint. This implies $$m^*(A \cup B) = m^*(A\setminus B \cup B\setminus A \cup A\cap B) = m^*(A\setminus B) + m^*(B \setminus A) + m^*(A\cap B).$$ It then follows that $$m^*(A\setminus B) + m^*(A \cap B) + m^*(B \setminus A) + m^*(A\cap B) = m^*(A \cup B) + m^*(A \cap B).$$ Therefore, we conclude that, for $A$ and $B$ measurable, $$m^*(A) + m^*(B) = m^*(A \cup B) + m^*(A \cap B).$$