Example
A discontinuous function can be Darboux integrable. For example take $$ \begin{align*} f(x) = \begin{cases} 1, & \text{if } x \in [0,1/2), \\ 2, & \text{if } x \in [1/2,1]. \end{cases} \end{align*} $$

Strategy

Note that \(f\) is discontinuous at \(\frac{1}{2}\). To show that \(f\) is Darboux integrable, then let \(\epsilon > 0\) be given. We want to show that there exists a partition \(P\) such that

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

Note now that if the interval is chosen to be entirely in \([0,1/2)\), then \(M_j = m_j = 1\). Similarly, if the interval is in \([1/2,1]\), then \(M_j = m_j = 2\). But if the interval was something like \((1/4,3/4)\), then \(M_j = 2\) and \(m_j=1\) so the difference is not small. Therefore, we want the interval containing \(1/2\) to be really small! Note now that if we pick the partitions \(\{0,x_1,x_2,1\}\) such that \(x_1 = \frac{1}{2} - \frac{\epsilon}{2}\) and \(x_2 = \frac{1}{2} + \frac{\epsilon}{2}\). Then, we’ll get the following intervals:

  • \([0,x_1]\): \(M_j = m_j = 1\)
  • \([x_1,x_2]\): Even though in this interval we have \(M_j = 2\) and \(m_j = 1\), its length is \(\frac{1}{2} + \frac{\epsilon}{2} - \left(\frac{1}{2} - \frac{\epsilon}{2}\right) = \epsilon\). Hence, this interval won't matter.
  • \([x_2,1]\): \(M_j = m_j = 2\)

Observe now that if we calculate \(U(f,P)\) and \(L(f,P)\), then we get the sum to be \(\epsilon\). Hence the function will be Darboux Integrable.

Proof

Let \(\epsilon > 0\) be given. choose the partition

$$ \begin{align*} P = \{ 0, x_1 = \frac{1}{2} - \frac{\epsilon}{2}, x_2 = \frac{1}{2} + \frac{\epsilon}{2}, 1 \} \end{align*} $$

Then, \(x_2 - x_1 = \epsilon\).

$$ \begin{align} U(f,P) &= \sum_{j=1}^{n}M_j(x_{j}-x_{j-1}) = 1 \cdot (x_1-0) + 2 \cdot \epsilon + 2 \cdot ( 1 - x_2) \\ L(f,P) &= \sum_{j=1}^{n}m_j(x_{j}-x_{j-1}) = 1 \cdot (x_1-0) + 1 \cdot \epsilon + 2 \cdot ( 1 - x_2) \end{align} $$

This means that

$$ \begin{align} U(f,P) - L(f,P) = 2\epsilon - \epsilon = \epsilon \end{align} $$

Therefore, \(f\) is Darboux integrable. \(\ \blacksquare\)

References