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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!

user136038
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  • Take a look at this to help with your formatting. Unfortunately at the moment I can't even understand what it is you're asking. – ah11950 Mar 26 '14 at 23:50
  • Both of these come about by applying property M2 to well-chosen $A$ and $B$. Try playing around with that a bit. – Santiago Canez Mar 26 '14 at 23:56
  • Apply English! Think of $m$ as doling out some thing to each set; M2 means amount doled out to disjoint sets is the sum doled out to individual sets; Now when you have $A\subset B$ split $B$ as $A$ and the part of $A$ omitted in $B$, and apply M2. – P Vanchinathan Mar 27 '14 at 01:01

1 Answers1

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(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., $$m(A)\leq m(B). $$

ZHN
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