Theorem 0.3
Suppose that \(f\) be Riemann integrable on \([a,b]\). Then, \(|f|\) is Riemann integrable.

Proof

Let \(f\) be Riemann integrable on \([a,b]\). Recall the reverse triangle inequality is

$$ \begin{align*} \bigl||a| - |b|\bigr| \le |a-b| \end{align*} $$

Now, for any subinterval \([x_{j-1},x_j]\), and for any \(x,y \in [x_{j-1},x_j]\), we have

$$ \begin{align*} \bigl||f(x)| - |f(y)|\bigr| \le |f(x)-f(y)| \quad\quad\quad (1) \end{align*} $$

We know \(m_j(f) \le f(x)\) for all \(x\) and \(f(y) \le M_j(f)\) for all \(y\) in this interval. Hence any difference \(|f(x)-f(y)|\) will be at most

$$ \begin{align*} |f(x)-f(y)| \le M_j(f) - m_j(f) \quad\quad\quad (2) \end{align*} $$

Combining (1) and (2)

$$ \begin{align*} \bigl||f(x)| - |f(y)|\bigr| \le M_j(f) - m_j(f) \quad\quad\quad (3) \end{align*} $$

Now, recall that \(A\) is a set of points, then

$$ \begin{align*} \sup_{a,b \in A} |a -b| = \sup A - \inf A \end{align*} $$

Using this fact, then taking the supremum over all \(x,y \in [x_{j-1},x_j]\), we obtain

$$ \begin{align*} M_j(|f|) - m_j(|f|) = \sup_{x,y} \bigl||f(x)| - |f(y)|\bigr| \quad\quad\quad (4) \end{align*} $$

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Then, for any partition \(\mathcal{P}\), we have

$$ \begin{align*} U(|f|, \mathcal{P}) - L(|f|, \mathcal{P}) \leq U(f, \mathcal{P}) - L(f, \mathcal{P}) \quad\quad\quad (6) \end{align*} $$

But we know that \(f\) is Riemann integrable, then

$$ \begin{align*} U(f, \mathcal{P}) - L(f, \mathcal{P}) < \epsilon \quad\quad\quad (7) \end{align*} $$

Combining (6) and (7) we get that \(|f|\) is Riemann Integrable:

$$ \begin{align*} U(|f|, \mathcal{P}) - L(|f|, \mathcal{P}) < \epsilon \end{align*} $$

Next, by the definition of absolute value

$$ \begin{align} -|f(x)| \leq f(x) \leq |f(x)| \end{align} $$

Integrating this, we get

$$ \begin{align} -\int_a^b |f(x)|dx \leq \int_a^b f(x)dx \leq \int_a^b |f(x)|dx \end{align} $$

References