Set function $\mu$ additive or not?

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$C:=\{(a,b] \times (0,1]:0 \le a \le b \le1 \}\cup \{(0,1]\times (a,b]:0 \le a \le b\le 1\}$

Set-function $\mu$: $$\mu((a,b] \times (0,1])=\mu((0,1] \times (a,b])=b-a$$ on $\Omega=(0,1]^2$

I need to show if $C$ is a semiring, which it isn't if I am correct, and if $\mu$ is additive.

First of all I don't know if I have to show that $\mu(\bigcup_{i=1}^mA_i)=\sum_{i=1}^m\mu(A_i)$ for disjunct $A_1,...A_m \in C$ or only $\mu(A\cup B)=\mu(A)+\mu(B)$ for disjunct $A,B \in C$?

My try was that $\mu$ is additive because 2 disjunct sets are either both in $\{(a,b] \times (0,1]:0 \le a \le b \le1 \}$ or both in $\{(0,1]\times (a,b]:0 \le a \le b\le 1\}$.