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Suppose I have a function $|1_A f|$, is it possible to take the $1_A$ out of the absolute value in an integral. I.e. is this correct? $$\int_\Omega |1_A f| d\mu=\int_\Omega 1_A |f|d\mu=\int_A|f| d\mu\leq \int_\Omega |f| d\mu$$

This seems obviously true as $1_A$ is just 0 or 1 but im not positive.

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Yes, this is correct. This is because $|1_A f| = 1_A |f|$. To see this, consider the case when $\omega \in A$, so that both sides are equal to $|f|$. Otherwise, both sides are equal to $0$.

More generally, recall that $|xy|=|x||y|$ when $x$ and $y$ are real, and that $|x|=x$ whenever $x \ge 0$.

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  • $\begingroup$ Yea I was convinced with the first the first line alone $\endgroup$ – badatmath Sep 11 at 16:20

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