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If I know that$\ X < \sup_iX_i$ for example and if $\ X_i $ and $X$ are integrable then when I integrate both sides, where do I put the integral on the RHS? I know that in some situations you can move the supremum inside and outside according to beppo levi etc but what I want to know is forgetting that, where is it always defined or where should I always put it to start off with before thinking about beppo levi or fatou or anything else?

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What does the inequality mean? Is it pointwise? On a set of full measure? Is the family of functions on the right hand side countable? –  Michael Greinecker Jun 7 '12 at 21:16
    
Sorry, So lets say instead, $\ X_i i=1,2,3... $ is a sequence of integrable functions. And X is an integrable function, all on $(\Omega,F,P)$ a measure space . If $\ X\leq Sup_iX_i.. $ –  Rosie Jun 7 '12 at 21:22
    
The supremum of a countable family of measurable functions is measurable. so you can put $\int X\leq\int\sup X_i$. Whether the integra preserves the strict inequality depends on what "<" means. If $X(\omega)<\sup_i X_i(\omega)$ for almost all $\omega$, you can write $\int X<\int\sup_i X_i$, –  Michael Greinecker Jun 7 '12 at 21:26
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If $\displaystyle X < \sup_i X_i$ then $\displaystyle \int X \le \int \sup_i X_i$.

You can't say "$<$", but only "$\le$"; in some cases they are equal.

And there are some simple cases where $\displaystyle\sup_i \int X_i$ is not the same as $\displaystyle \int \sup_i X_i$.

For example, suppose $$ X_1(x) = \begin{cases} 1 & \text{if }0\le x \le 1/2, \\ 0 & \text{if } 1/2<x\le 1, \end{cases} $$ and $X_2 = 1 - X_1$.

Then $$ \sup_{i\in\{1,2\}} \int_0^1 X_i(x)\,dx = \frac 1 2 $$ and $$ \int \sup_{i\in\{1,2\}} X_i(x)\,dx = 1. $$

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