# Taking indicator function out of absolute value

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.

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 $$x \ge 0$$.