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My question is somehow similar to these equations in here, in here, and in here

let $\mu_1$ and $\mu_2$ be the two measures on $\mathbb R$ such that $\mu_1((a,b))= \mu_2((a,b))$ where $a,b \in \mathbb{R}$ for which $\mu_{1}(\{ a\})=0$, $\mu_{1}(\{ b\})=0$, $\mu_{2}(\{ a\})=0$, and $\mu_{2}(\{ b\})=0$. Show that $\mu_1$ and $\mu_2$ are equal.

I am not sure how would I use $\pi$-system and $\lambda$-system in here. Any hints are appreciated.

Edit: In the question, it was not mentioned if the measures were finite, but it seems to be a necessary assumption. So, assume it is finite i.e $\mu_1((a,b))= \mu_2((a,b)) < \infty$

user919322
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    Not true. For example, $\mu_1$ the counting measure except it is $0$ at $0$, $\mu_2$ similar except it is $0$ at $1$. There are no $a,b$ for which $\mu_i({a})=0$ (or $\mu_i({b})=0$) so the condition "$\mu_1((a,b))=\mu_2((a,b))$ for all $-\infty<a<b<\infty$ for which $\mu_1({a})=\mu_2({a})=\mu_1({b})=\mu_2({b})=0$" is vacuously true. $\mu_1,\mu_2$ are measures on $\mathbb{R}$, but $\mu_1({0})=0\neq1=\mu_2({0})$, for example. – user10354138 May 01 '21 at 14:57
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    Have you omitted an assumption? If not, then I don't think this is true: take $\mu_1$ to be the counting measure and $\mu_2=2\mu_1$. Then the assumption is vacuously true, because $\mu_i({x})$ is never $0$. – Rob Arthan May 01 '21 at 15:04
  • @user10354138 I have written it now as it is in my notes. – user919322 May 01 '21 at 15:12
  • @user919322 You are missing an assumption here. In the title you said finite measures, but in the question this assumption is never stated. For finite measures $\mu_1,\mu_2$ the conclusion holds (there are only countably many atoms which you can immediately deal with and the atomless case is easy to prove), but for $\mathbb{R}\cup{\infty}$-valued measures this fails as both my and Rob's examples shows. – user10354138 May 01 '21 at 15:26
  • @user10354138 I have edited the question. Thank you! – user919322 May 01 '21 at 19:23

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