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 $G^{B}((a,b])=G^{L}((a,b])$.

  • $\begingroup$ Thank you! I edited the question, sorry. What I should have writtten actually was whether we can assert for every half-closed intervals. $\endgroup$ – Caio Lorecchio Apr 2 at 0:06
  • $\begingroup$ @CaioLorecchio I have edited the answer accordingly. $\endgroup$ – Kavi Rama Murthy Apr 2 at 0:18
  • $\begingroup$ $G^{H}(B)=1/8$ and $G^{L}(B)=1/2$, so the measures do not coincide. Am I missing something here? $\endgroup$ – Caio Lorecchio Apr 2 at 0:52
  • $\begingroup$ @CaioLorecchio Sorry, that example was wrong. $\endgroup$ – Kavi Rama Murthy Apr 2 at 5:34
  • $\begingroup$ @CaioLorecchio I have revised my answer. $\endgroup$ – Kavi Rama Murthy Apr 2 at 7:32

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