Given that a function $g(x)$ is a monotone increasing function. Its domain is an interval [c,d]. There are two sets $a_j \in A, j= 1,...,J$ and $b_k \in B,k=1,...,K$ on this interval, which satisfy $c_z \in C = B$\A and $c_z \geq a_j$ for all $a_j \in A$. For example, the interval $[c,d]=[0,8]$,$A=\{0,1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5,6,7\}$. In this case, does the following inequality hold? If yes, could you give me some hints to prove that?
$$\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)$$
No. Take $g(x)=x$, $[c,d]=[0,1]$, $A=\{1\}$ and $B=\{0,1\}$ then $$\frac{1}{2}=\frac{1}{K} \sum_{k=1}^{K}g(b_k) < \frac{1}{J} \sum_{j=1}^{J}g(a_j)=1.$$
On the other hand, if you assume that if $b\in B\setminus A$ then $b>\mu:=\frac{1}{|A|}\sum_{a\in A}g(a)$, it follows $$\frac{1}{|B|} \sum_{b\in B}g(b)=\frac{1}{|B|} \sum_{a\in A}g(a)+\frac{1}{|B|} \sum_{b\in B\setminus A}g(b)\\ >\frac{|A|}{|B|} \cdot \mu+\frac{1}{|B|} \sum_{b\in B\setminus A}\mu =\frac{|A|}{|B|} \cdot \mu+\frac{|B|-|A|}{|B|} \cdot \mu=\frac{1}{|A|}\sum_{a\in A}g(a).$$