Suppose we are given two probability measures $\mathbb{G}^{H}$ and $\mathbb{G}^{L}$ with same support $\text{supp}(\mathbb{G})$.
Suppose as well that there exists an integrable function $\gamma$ such that $\mathbb{G}^{H}(B)=\int_{B}\gamma(r)\mathbb{G}^{L}(dr)$ for every subset $B\in \mathcal{B}$, the Borel $\sigma$-algebra of $[0,1]$. This function is a.s. positive, finite and different than 1. It is also increasing.
Question: Can we assert that $|\mathbb{G}^{H}(B)-\mathbb{G}^{L}(B)|>0$ for any half-closed interval $B=(a,b]$ such that $0<\mathbb{G}^{L}(B)<1$?
Let $G^{L}$ b e Lebesgue measure on $[0,1]$ and $G^{B}(E)=\int_E 3x^{2} dx$. Then the condition $G^{B}((a,b])=G^{L}((a,b])$ reduces to $b^{3}-a^{3}=b-a$. There are plenty of intervals with this property; in fact for any $a\in (0,\frac 1 {\sqrt 3})$ we can find $b \in (a,1)$ such that $a^{2}+b^{2}+ab=1$ which implies $G^{B}((a,b])=G^{L}((a,b])$.