$|f| $ is Lebesgue integrable , does it implies $f$ is also?

Question

If $ f $ is Lebesgue integrable then $|f|$ is Lebesgue integrable but does the converse of the result is also true?

Answer 1

Assume $f$ to be measurable.

$f$ is bounded by $|f|$ and so it is $L^1$, and hence Lebesgue integrable.

On the other hand, $f$ is lebesgue iff $\int_{X}f^+ \, d\mu<\infty$ and $\int_{X}f^- \, d\mu<\infty$, but we know that $|f|=|f^+ + f^-| \leq |f^+|+|f^-|$ and so $|f|$ is integrable if $|f|$ is.

Answer 2

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.

Answer 3

Think of |f| as a division of f into two functions: $f_+$ and $f_-$.

$f_+$ we define as equal to f on the domain {x: f(x) is non-negative}, and 0 on all other x.

$f_-$ we define as equal to -f on the domain {x: f(x) is negative} and 0 elsewhere. $|f|=f_+ + f_-$.

If |f| is finite, then necessarily both $f_+$ and $f_-$ are finite.