Let $A,B \subset \mathbb R$ with $m^*(A)=0$. Prove $m^*(A \cup B)=m^*(B).$ Let $A,B \subset \mathbb R$ with $m^*(A)=0$. Prove $m^*(A \cup B)=m^*(B).$

$m^*(A \cup B)=m^*(A) + m^*(B)$ if A and B are disjoint. 

In general, $$m^*(A \cup B) \leq m^*(A) + m^*(B)=0+m^*(B)=m^*(B)$$ but $m^*(A \cup B) &lt; m^*(B)$ is impossible so $m^*(A \cup B)=m^*(B).$

Is that valid or am I missing something?
First, since $m^*(A) =0$. Then, $$m^*(A \cup B) \leq m^*(A) + m^*(B) \leq m^*(B).$$
It suffices to show that $m^*(B) \leq m^*(A \cup B)$. We need to consider three cases. 



If $A = \emptyset$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.



If $\emptyset \neq A \subseteq B$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.



If $\emptyset \neq A \nsubseteq  B$, then $B \subset A \cup B$ and $m^*(B) \leq m^*(A \cup B)$.



As you can see all the three cases gives $m^*(B) \leq m^*(A \cup B)$ as required.
First, since $m^*(A) =0$. Then, $$m^*(A \cup B) \leq m^*(A) + m^*(B) \leq m^*(B).$$
It suffices to show that $m^*(B) \leq m^*(A \cup B)$. We need to consider three cases. 



If $A = \emptyset$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.



If $\emptyset \neq A \subseteq B$, then $A \cup B = B$ and $m^*(B) \leq m^*(A \cup B) = m^*(B)$.



If $\emptyset \neq A \nsubseteq  B$, then $B \subset A \cup B$ and $m^*(B) \leq m^*(A \cup B)$.



As you can see all the three cases gives $m^*(B) \leq m^*(A \cup B)$ as required.