If $f$ is a Riemann integrable prove $|f|$ is also Riemann integrable 
Show that if $f$  is Riemann integrable on $[a,b]$ then $|f|$ is also Riemann integrable on $[a,b]$.

My idea is:
let $f$ be in $[a,b]$ less than $|f|$, since $f$ is integrable then $|f|$ is also integrable on $[a,b]$.
 A: Here is a fancier approach. A bounded function on a bounded interval is Riemann integrable if and only if it is continuous everywhere except on a set of Lebesgue measure zero.  
Suppose that $f$ is Riemann integrable and $g$ is continuous on an interval containing the range of $f$.  Then  $g\circ f$ is continuous wherever $f$ is
continuous; therefore, $g\circ f$ is continuous everywhere but on a set of Lebesgue measure zero.  We conclude that $g\circ f$ is Riemann integrable.  
This specializes immediately to your case.
A: More generally, one can prove that if $f: [a,b] \rightarrow [c,d]$ is Darboux integrable and $\varphi: [c,d] \rightarrow \mathbb{R}$ is continuous, then the composite function $\varphi \circ f: [a,b] \rightarrow \mathbb{R}$ is Darboux integrable.  (Then take $\varphi(x) = |x|$ to answer the OP's question.)  A proof appears on page 7 of these notes (Wayback Machine).  (It comes directly from Russell Gordon's text Real Analysis: A First Course, which was the text for the class from which these notes were made.)
Comments:

*

*Note that I said "Darboux integrable": this is the version of the integral which uses upper and lower sums instead of Riemann sums.  As I say in my notes, it is (so far as I know...) not so easy to adapt this argument to show Riemann integrability.


*But perhaps the OP actually means Darboux integrability!  It is common to use the term "Riemann integral" to refer to either the Darboux or the Riemann integral.  This is not a horrible mistake, because there is a theorem that says that these two integrals are equivalent (i.e., a function is Riemann integrable iff it is Darboux integrable, and if so these two integrals have the same value), but I think it is sloppy in a way which can  cause confusion for students.


*One advantage of this general approach is that it leads to an easy proof that the product of Darboux integrable functions is Darboux integrable.


*The result does not work the other way around: if $f$ is Riemann/Darboux integrable and $\varphi$ is continuous, then the composition $f \circ \varphi$ (when defined) need not be Riemann/Darboux integrable.  In my notes I state this but mention that the examples I know are too hard to include.  (And now of course I would have to think about constructing such examples from scratch, because I didn't write them down!)  Does perchance anyone know of a nice, simple example of this?
A: Hints:
Note that $f$ and thus $|f|$ is bounded on $[a,b]$.
First show that
for any interval $I=[c,d]\subset[a,b]$,
$$\tag{1}
\sup_{x\in I} |f(x) | -\inf_{x\in I}|f(x)| \le \sup_{x\in I}  f(x)  -\inf_{x\in I} f(x) 
$$
Next show that inequality (1) implies that for any partition $P$ of $[a,b]$:
the upper Riemann sums $U(|f|, P)$  and $U(f, P)$  and the lower Riemann sums
$L(|f|, P)$  and $L(f, P)$ satisfy
$$
U(|f|, P)-L(|f|, P) \le U(f, P)-L(f, P).
$$
Then argue that $U(|f|, P)-L(|f|, P) $ can be made as small as desired by taking an appropriate partition $P$ of $[a,b]$.
Finally conclude that $|f|$ is integrable.



Per kahen's comment (see his link):

In the above, the following characterization of Riemann integrability is used:
For a partition $P=\{x_0,x_1,\ldots, x_n\}$ of $[a,b]$, define the upper Riemann sum of the bounded function $f$ by:
$$
U(f,P)= \sum_{j=1}^n \bigr( \sup_{x\in[x_{j-1},x_j]} f(x)\bigl)(x_j-x_{j-1});
$$
and
the lower Riemann sum   by:
$$
U(f,P)= \sum_{j=1}^n \bigr( \inf_{x\in[x_{j-1},x_j]} f(x)\bigl)(x_j-x_{j-1})
$$
Then $f$ is Riemann integrable on $[a,b]$ if and only if $f$ is bounded on $[a,b]$, and for every $\epsilon>0$  there is a  partition $P$ such that $U(f,P)-L(f,P)<\epsilon$.

