Where $f$ is a $\mu$-integrable function from a measure space $X$ to $\mathbb{R}$. I am reading through the Radon-Nikodym theorem in Folland's Real Analysis and I am not sure how to prove this point. This is on pg 90 of Folland. $\nu$ is a $\sigma$-finite signed measure and $\mu$ is a $\sigma$-finite measure. I understand why this is true intuitively.
Why is a measure $\nu = \int f d \mu$ absolutely continuous with respect to $\mu$?
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real-analysis
measure-theory
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0Shouldn't this be under the assumption that $\mu(E)=0$ implies $\nu(E)=0$ for every set $E$? – 2017-01-14
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0That is the definition of absolutely continuous. My question is how to prove that is true. – 2017-01-14
2 Answers
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Suppose $\mu (E)=0,\ $and assume wlog that $f\ge 0$. Now, choose an increasing sequence of simple functions, $f_n,$ such that $f_n\to f.$ Then, applying the MCT, and the definition of the integral for simple functions, we have $\nu (E)=\int_Efd\mu=\int_E (\lim f_n)d\mu =\lim \int_Ef_nd\mu =0.$
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You just have to prove that if $f\in L^1 (\mu)$ and $\mu(E) = 0$, then $\int_E f d\mu = 0$. It suffices to prove this for $f\ge 0$. First prove it for simple functions, then approximate $f$ by a sequence of simple functions.