TITLE: $|f| $ is Lebesgue integrable , does it implies $f$ is also?
QUESTION [6 upvotes]: If $ f $ is Lebesgue integrable then $|f|$ is Lebesgue integrable but does the converse of the result is also true?

REPLY [6 votes]: No, it does not: If $E$ is (bounded and) non-measurable, then $f = 2\chi_{E} - 1$ is everywhere equal to $\pm1$, so $|f| \equiv 1$, but $f$ is non-measurable, hence not integrable.