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Let $g_n=\max(f_1,f_2,...,f_n)$ where $f_i:X \to [0,\infty]$ are measurable functions on a measure space $X$. Let $\nu$ and $\mu$ be measures. I am trying to show that $\int_E g_n d\mu \leq \nu(E)$ if for each $f_i$, $\int_E f_i \leq \nu(E)$ for all measurable $E\subseteq X$. This is for a proof of the Radon-Nikodym theorem on pg 90 of Folland. My idea (based off some hints in Folland) is to partition $E$ into sets where $g_n(x)=f_i(x)$ for some specific $i$ let these sets be $E_i$. Then, $$\int_E g_n = \sum_{i=1}^{n}\int_{E_i} f_i \leq \sum_{i=1}^{n} \nu(E_i) = \nu(E)$$ The problem is that I have to show the $E_i$ are measurable in order to use both the inequality and the last equality. So my question is either how can I show they are measurable or if not is there another way to do this.

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It suffices to look at the case $n=2$. Let $E$ be measurable and let $E_1 = \{x\in E : f_1(x) \ge f_2(x)\}$ and $E_2 = E - E_1$. The two sets are clearly measurable, and $\int_E g = \int_{E_1} f_1 + \int_{E_2}f_2 \le \nu(E_1) + \nu(E_2) = \nu(E)$.

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  • $\begingroup$ Why is $E_1$ measurable. Also why does it suffice to consider $n=2$? Do you use induction? $\endgroup$ – edenstar Jan 15 '17 at 2:48
  • $\begingroup$ @edenstar because $E_1 = E \cap (f_1 - f_2)^{-1}([0,\infty))$. Yes, use induction $\endgroup$ – user384138 Jan 15 '17 at 2:55
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Since $f_i$ are measurable, so are $f_i - f_j$, and $L_{ij} = \{x: f_i(x) \le f_j(x)\} = \{x: (f_i - f_j)(x) \le 0\}$ is measurable. Your sets $E_i$ can be built up from those using Booolean operations.

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  • $\begingroup$ I'm not sure why the $L_{ij}$ are measurable. $\endgroup$ – edenstar Jan 15 '17 at 2:50
  • $\begingroup$ Once you know $f_i - f_j$ is measurable, $\{x: (f_i - f_j)(x) \in E\}$ is measurable for any Borel set $E$, by definition of measurable function. $\endgroup$ – Robert Israel Jan 15 '17 at 7:31

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