Difference between absolutely continuous measures

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 $$0<\mathbb{G}^{L}(B)<1$$?

1 Answer

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

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