## Appendix A  
**Proof of Properties of Theorem 1. ** The four properties in Definition 2 are proved as follows. (1) From Property 1, for \(\forall x\in U,0\leq 1-\frac{1}{2}(\left|\mu_{A}^{2}(x)-\mu_{B}^{2}(x)\right|+ \left|\
u_{A}^{2}(x)-\
u_{B}^{2}(x)\right|+\left|\pi_{A}^{2}(x)\right.\)  
\(-\pi_{B}^{2}(x)\right|)\leq 1\). Therefore, \(0\leq S(A,B)\leq 1\). (2) \(S(A,B)=1\) if and only if \(\mu_{A}(x)=\mu_{B}(x)\), \(\
u_{A}(x)=\
u_{B}(x)\) and \(\pi_{A}(x)=\pi_{B}(x)\). For \(\forall x\in U\), if and only if \(A=B\). Therefore, \(S(A,B)=1\), if and only if \(A=B\). (3) According to Equation (2), \(S(A,B)=S(B,A)\) is obvious. (4) If \(A\subseteq B\subseteq C\), then for \(\forall x\in U,\mu_{A}(x)\leq\mu_{B}(x)\leq\mu_{C}(x)\) and \(\
u_{A}(x)\geq\
u_{B}(x)\geq\
u_{C}(x)\). Assume that \(Q(A,C)(x)=\left|\mu_{A}^{2}(x)-\mu_{C}^{2}(x)\right|+\left|\
u_{A}^{2}(x)-\
u_ {C}^{2}(x)\right|+\left|\pi_{A}^{2}(x)-\pi_{C}^{2}(x)\right|\), and \(Q(A,B)(x)=\left|\mu_{A}^{2}(x)-\mu_{B}^{2}(x)\right|+\left|\
u_{A}^{2}(x)-\
u_ {B}^{2}(x)\right|+\left|\pi_{A}^{2}(x)-\pi_{B}^{2}(x)\right|\). For \(\forall x\in U,\)\(Q(A,C)(x)\geq Q(A,B)(x)\). Next, we will prove this from the following six cases: 1 \(\pi_{A}(x)\leq\pi_{B}(x)\leq\pi_{C}(x)\); 2 \(\pi_{A}(x)\leq\pi_{C}(x)<\pi_{B}(x)\); 3 \(\pi_{B}(x)<\pi_{A}(x)\leq\pi_{C}(x)\); 4 \(\pi_{B}(x)\leq\pi_{C}(x)<\pi_{A}(x)\); 5\(\pi_{C}(x)<\pi_{A}(x)\leq\pi_{B}(x)\); and 6\(\pi_{C}(x)<\pi_{B}(x)<\pi_{A}(x)\). **Case 1. ** If \(\pi_{A}(x)\leq\pi_{B}(x)\leq\pi_{C}(x)\), \(x\in U\), then \(Q(A,C)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{C}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{B}^{2}(x)\big{)}\). Obviously, \(Q(A,C)(x)\geq Q(A,B)(x)\). **Case 2. ** If \(\pi_{A}(x)\leq\pi_{C}(x)<\pi_{B}(x)\), then \(Q(A,C)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{C}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{B}^{2}(x)\big{)}\). Therefore, \(Q(A,C)(x)\geq Q(A,B)(x)\). **Case 3. ** If \(\pi_{B}(x)<\pi_{A}(x)\leq\pi_{C}(x)\), then \(Q(A,C)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{C}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\
u_{B}^{2}(x)-\
u_{A}^{2}(x)\big{)}\). We can prove this by reducing to absurdity: \(Q(A,C)(x)\geq Q(A,B)(x)\). Assume that \(Q(A,C)(x)<Q(A,B)(x)\), i.e., \(\
u_{A}^{2}(x)-\
u_{C}^{2}(x)<\mu_{B}^{2}(x)-\mu_{A}^{2}(x)\), then \(\mu_{A}^{2}(x)+\
u_{A}^{2}(x)<\mu_{B}^{2}(x)+\
u_{C}^{2}(x)\leq\mu_{C}^{2}(x)+ \
u_{C}^{2}(x)\). So, \(1-\pi_{A}^{2}(x)<1-\pi_{C}^{2}(x)\). Therefore, \(\pi_{A}(x)>\pi_{C}(x)\). This contradicts \(\pi_{A}(x)\leq\pi_{C}(x)\). Therefore, the assumption \(Q(A,C)(x)<Q(A,B)(x)\) does not hold. Therefore, \(Q(A,C)(x)\geq Q(A,B)(x)\). **Case 4. ** If \(\pi_{B}(x)\leq\pi_{C}(x)<\pi_{A}(x)\), then \(Q(A,C)(x)=2\big{(}\mu_{C}^{2}(x)-\mu_{A}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\mu_{B}^{2}(x)-\mu_{A}^{2}(x)\big{)}\). We can obtain \(Q(A,C)(x)\geq Q(A,B)(x)\). **Case 5. ** If \(\pi_{C}(x)<\pi_{A}(x)\leq\pi_{B}(x)\), then \(Q(A,C)(x)=2\big{(}\mu_{C}^{2}(x)-\mu_{A}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\
u_{A}^{2}(x)-\
u_{B}^{2}(x)\big{)}\). We can prove this by reducing to absurdity: \(Q(A,C)(x)\geq Q(A,B)(x)\). Assume that \(Q(A,C)(x)<Q(A,B)(x)\), i.e., \(\mu_{C}^{2}(x)-\mu_{A}^{2}(x)<\
u_{A}^{2}(x)-\
u_{B}^{2}(x)\), then \(\mu_{C}^{2}(x)+\
u_{C}^{2}(x)\leq\mu_{C}^{2}(x)+\
u_{B}^{2}(x)<\mu_{A}^{2}(x)+ \
u_{A}^{2}(x)\), namely, \(1-\pi_{C}^{2}(x)<1-\pi_{A}^{2}(x)\). So, we know that \(\pi_{A}(x)<\pi_{C}(x)\) contradicts \(\pi_{A}(x)>\pi_{C}(x)\). Therefore, the assumption \(Q(A,C)(x)<Q(A,B)(x)\) does not hold. Therefore, \(Q(A,C)(x)\geq Q(A,B)(x)\). **Case 6. ** If \(\pi_{C}(x)<\pi_{B}(x)<\pi_{A}(x)\), then \(Q(A,C)(x)=2\big{(}\mu_{C}^{2}(x)-\mu_{A}^{2}(x)\big{)}\) and \(Q(A,B)(x)=2\big{(}\mu_{B}^{2}(x)-\mu_{A}^{2}(x)\big{)}\). Therefore, \Similarly, \(S(A,C)\leq S(B,C)\). Therefore, if \(A\subseteq B\subseteq C\), then \(S(A,C)\leq\min\{S(A,B),S(B,C)\}\). \(\square\)  
**Proof of Properties of Theorem 2. ** The divergence measure should satisfy the four properties in Definition 3. The proofs are as follows. (1) Obviously, \(D(A,B)=D(B,A)\). (2) \(D(A,A)=0\).