Proving $f$ is Lebesgue integrable iff $|f|$ is Lebesgue integrable. How can we prove the statement? 
$f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable. 
First, suppose $f$ is Lebesgue integrable, by definition, $f$ is measurable and nonnegative, so $f=|f|$ and $\displaystyle \int_E f \, d \mu=\int_E|f| \, d\mu$ implies that $|f|$ is Lebesgue integrable. 
Second, suppose $|f|$ is Lebesgue integrable, if $f$ is nonnegative, then $f=|f|$, so $f$ is Lebesgue integrable; if $f$ is negative, then how can I prove? 
 A: 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.
A: 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 $|f|=f^++f^-$.
