Let $\mu:\mathscr{C}\to\mathbb{R}=[-\infty,+\infty]$ is monotonous if $\mu(\emptyset)=0$ adnd for $E,F\in\mathscr{C}$, $E\subset F$ implies that $\mu(E)\leqslant\mu(F)$. If $\mathscr{C}$ is a ring, prove that a non-negative additive set function is monotonous.
I have devised a proof that was wrong but I do not know the reason for that.
Let $A,B\in\mathscr{C}$ then as $\mathscr{C}$ is a ring then $A\cup B\in\mathscr{C}$. Taking $H=A\cup B$ then as the measure is non-negative $\mu(A)>0$ and $\mu(B)>0$. Then it is obviously seen that:
$A\subset(A\cup B)=H$ and $\mu(A)\leqslant \mu(A\cup B)=\mu(A)+\mu(B)=mu(H)$
As $A,B$ are arbitrary sets of $\mathscr{C}$ we conclude $\mu$ is monotonous.
Questions:
Why is my proof wrong? Which are the alternatives?
Thanks in advance!
For $E,F\in \mathscr{C}$ s.t. $E\subset F$, $$ \mu(F)=\mu(E)+\mu(F\setminus E)\ge \mu(E). $$