# $\lambda$ is absolutely continuous with respect to $\mu$ then $\lambda (A) -\int_A \epsilon I_E d\mu \geq 0$

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?

• do you mean absolute continuous instead of absolute convergent`? – Cettt Jan 7 '17 at 10:42
• Yes. I was thinking about another question which I should show that the integral is absolutely convergent, so I made the mistake. – Zohreh Aliabadi Student Jan 7 '17 at 10:45