Q:

A sufficient condition to ensure $g(t)=a \ln t+b$

Let $(X,\mu)$ be a measure space, $\mu(X)=1$, and for any measurable $A\subset X$, there exists  measurable subset $B\subset A$, such that $0<\mu(B)<\mu(A)$. 
$g(t)$ is a continuous function on $[0,\infty)$. If 
$$g(\exp(\int_X \ln f d \mu))=\int_X g(f)d \mu.$$
for any non-negative measurable function $f$, show that $\exists\ a,b\in\Bbb R$, such that $g(t)=at+b$. 
Can we show that $a,b\geq 0$? I do not think so. But for the more general result, I have no idea.

A:

As in  the title we can prove that $g(t)=a \ln t +b$. $a$ and $b$ can be any two real numbers. [$g(t)=at+b$ is wrong]. 
It is well known that a measure satisfying  the given hypothesis (which is called a non-atomic measure) assumes all values between  $0$ and $1$. [ This means for any $t \in [0,1]$ there exists a set $A$ such that $\mu (A)=t$]. Ref. Simpler proof - Non atomic measures
Let $f =I_B+c I_{B^{c}}$. Then, from the given equation we get $g (c^{t}) =g(1)(1-t)+g(c) t$ where $t=\mu (B^{c})$. Put $s=c^{t}$ to get $g(s)=g(1)(1-\frac {\ln s} {\ln c})+g(c){\frac {\ln s} {\ln c}}$  for all $s$ from  which the result follows. 

