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Let $(X,M,\mu)$ be a measure space, $\lambda$ and $\mu$ positive measures on $X$ which $\lambda$ is absolutely continuous with respect to $\mu$. Then either $\lambda=0$, or $\exists \; \epsilon >0$ and $E\in M$ with $\mu(E)\neq 0$ such that for all $A\in M$ $$\lambda (A) -\int_A \epsilon I_E d\mu \geq 0$$It is straightforward with the Radon-Nikodym theorem, but the question asks to give a proof without using that theorem. If $$\lambda \neq0$$
on the contrary suppose that $\forall \epsilon >0$ and for all $E \in M$ with $\mu(E)\neq 0$, $\lambda (A) -\int_A \epsilon I_E d\mu < 0$ holds for all $A \in M$. Since $\lambda$ is absolutely continuous with respect to $\mu$ then $\forall \; \epsilon >0$, $\exists \; \delta >0$ such that $\lambda(E)<\epsilon$ whenever $\mu (E)< \delta$. Then I tried to split the integral in case of $A=E \cup (A\setminus E)$. Any hint on how to do this question?

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  • $\begingroup$ do you mean absolute continuous instead of absolute convergent`? $\endgroup$ – Cettt Jan 7 '17 at 10:42
  • $\begingroup$ Yes. I was thinking about another question which I should show that the integral is absolutely convergent, so I made the mistake. $\endgroup$ – Zohreh Aliabadi Student Jan 7 '17 at 10:45
  • $\begingroup$ Consider the signed measure $\alpha_n=\lambda-(1/n)\mu$ for $n=1,2,3,....$ By Hahn decomposition, there exists a set $E_n$ in M so that $\alpha(A)\geq 0$ for every $A \in E_n$ and $\alpha(A)\leq 0$ for every A in $X\setminus E_n. $\endgroup$ – Zohreh Aliabadi Student Jan 9 '17 at 12:30
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Consider the signed measure $\alpha_n=\lambda-(1/n)\mu$ for $n=1,2,3,....$ By Hahn decomposition, there exists a set $E_n \in M$ so that $\alpha(A)\geq 0$ for every $A \in E_n$ and $\alpha(A)\leq 0$ for every $A \in X\setminus E_n$.

Suppose $\mu$ and $\nu$ are not trivially zero. Now $\mu(E_n)$ is not zero for at least one n. Because suppose $\mu(E_n)=0$ for every n, $let E_0=\cup_nE_n$. Then $\mu(E_0)=0$ and $n\lambda(A)\leq \mu(A)$ for every $A in X\setminus E_0$. There is at least one set $A \in X\setminus E_0$ so that $\lambda(A)\neq 0$. Then $\mu(A)=\infty$. Hence you get a contradiction. Set $E=E_n$, where $\mu(E_n)$ is not zero and let $\epsilon=1/2$. It is easy to verify that $\lambda(A)-(1/2)\int_A\chi_Ed\mu\geq 0$ for every $A \in M$.

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