Proving $f$ is Lebesgue integrable iff $|f|$ is Lebesgue integrable.

Solution 1:

That $f$ is measurable and $\int |f| \,d\mu < \infty$ is the definition of Lebesgue-integrability of $f$. (There's no need for $f$ to be non-negative. Rather, one first defines the value of the integral for non-negative functions and then one uses that to define the Lebesgue integral of functions whose ranges may include both negative and non-negative numbers.)

That $|f|$ is Lebesgue-integrable would therefore involve the absolute value of the absolute value of $f$. But that's the same as the absolute value of $f$.

The statement "$f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable." is not true unless there is an assumption that $f$ is measurable. A counterexample is that $f$ is $-1/2$ plus the indicator function of a non-measurable set in a space whose total measure is $1/2$. Then $|f|$ is integrable (and its integral is $1$) but $f$ is not measurable.

Solution 2:

Simply remember the definition of $\mathcal{L}^1$ space (in the case of $\Bbb R^n$ with the borelian $\sigma$-algebra; the abstract case is similar): given $(\Bbb R^n,\mathcal{B}(\Bbb R^n),m)$, where $m$ is the Lebesgue measure on the measurable space $(\Bbb R^n,\mathcal{B}(\Bbb R^n))$, then $$ \mathcal{L}^1(\Bbb R^n,\mathcal{B}(\Bbb R^n),m):=\{f:\Bbb R^n\to\Bbb R\;:\;f\;\;\mbox{is Lebesgue measurable and both}\;\int_{\Bbb R^n}f^+\,dm,\;\int_{\Bbb R^n}f^-\,dm\;\mbox{are finite}\}\;\;. $$ Now simply observe that $f=f^+-f^-$ and $|f|=f^++f^-$.