$f \in \mathscr{R}[a,b] \implies |f| \in \mathscr{R}[a,b]$

Question

If $f$ is Riemann integrable on $[a,b]$ then so is $|f|$

Here is my attempt:

Since $f \in \mathscr{R}[a,b]$, then $f$ is continuous almost everywhere on $[a,b]$ and is bounded. Let $V$ be a set of points where it is discontinuous so $V$ is of measure zero.

Let $S$ be the set of intervals where $f$ is negative. That is, $S=\{I_1, I_2,...,I_n\}$.

Let $T=\{x_0,x_1,...,x_p\}$ be the set of endpoints of these intervals.

Then $|f|$ is discontinous on $V \cup T$. Since $V \cup T$ is of measure zero, then $|f|$ is continuous almost everywhere on $[a,b]$. It is also bounded since $f$ is bounded.

Hence, $|f| \in \mathscr{R}[a,b]$

$\Box$

Is my proof correct? How can I improve it? What's another way to prove this?

Answer 1

$|f|$ is continuous wherever $f$ is continuous, because $|\cdot|$ is a continuous function.

Answer 2

Another proof.

It suffices to note that \begin{align*} \sup_{I}|f|-\inf_{I}|f|\leq\sup_{I}f-\inf_{I}f. \end{align*} The reason is like the following.

We have $\sup_{I}|f|-\inf_{I}|f|=\sup_{x,y\in I}(|f(x)|-|f(y)|)$. And $|f(x)|-|f(y)|\leq|f(x)-f(y)|\leq\sup_{I}f-\inf_{I}f$.

Note that we have $f(x)-f(y)\leq\sup_{I}f-\inf_{I}f$, and by symmetry, $f(y)-f(x)\leq\sup_{I}f-\inf_{I}f$.

Also note that $\sup_{x\in I}|f(x)|-\inf_{y\in I}|f(y)|=\sup_{x\in I}|f(x)|+\sup_{y\in I}(-|f(y)|)=\sup_{x,y\in I}(|f(x)|-|f(y)|)$. Note that $\sup(A+B)=\sup A+\sup B$.

Now we are to apply Cauchy criterion for Riemann integrability.