Let $S$ be a finite, non-empty set and $m:2^S\to R$ a function with the following properties
- $M1$: $\forall A\in2^S, m(A)\ge0$
- $M2$: $\forall A, B\in 2^S, A\cap B=\varnothing\Longrightarrow m(A\cup B)=m(A)+m(B)$
$(a)$ Prove that $m(\varnothing) = 0$.
$(b)$ Prove that $m$ is monotone, i.e., $\forall A,B \in 2^S, A\subseteq B\Longrightarrow m(A)\le m(B)$.
I'm just stuck on how to even start these problems. Any help? Thanks!
(a) As $\emptyset \cap \emptyset =\emptyset$, then $$m(\emptyset)=m(\emptyset \cup \emptyset)=m(\emptyset)+ m(\emptyset) ,$$ then, subtracting $m(\emptyset)$ in the equality above, we obtain $$0=m(\emptyset). $$
(b) $m$ is monotone. Indeed, if $A\subset B$, then $(B\setminus A)\cap A=\emptyset$ and the union $A\cup (B\setminus A)=B$ is disjoint, then, $$m(B)=m((B\setminus A)\cup A)=m(B\setminus A)+m(A)\geq m(A), $$ i.e., $$m(A)\leq m(B). $$