If $f$ is a Riemann integrable prove $|f|$ is also Riemann integrable

Question

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 $[a,b]$.

Answer 1

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.

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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 $U(f,P)-L(f,P)<\epsilon$.

Answer 2

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.

Answer 3

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:

1. 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.

2. 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.

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

4. 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?