Theorem
All continuous functions on \([a,b]\) are Darboux integrable.

Notes

To show that \(f\) is Darboux integrable, we need to show that for all \(\epsilon > 0\), there exists a partition \(P\) such that

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

If we expand this, then we want

$$ \begin{align} \sum_{j=1}^{n} M_j(x_j - x_{j-1}) - \sum_{j=1}^{n} m_j(x_j - x_{j-1}) = \sum_{j=1}^{n} (M_j-m_j)(x_j - x_{j-1}) < \epsilon \end{align} $$

We can choose the intervals to be really small but that doesn’t guarantee that \(M_j - m_j\) is small. In fact, \(M_j - m_j\) is guaranteed to be as small only when \(f\) is uniformly continuous. Hence, we are going to choose that \(x_j - x_{j-1} < \delta\) and this choice will force \(|f(x_j) - f(x_{j-1})| < \epsilon\). But there is something important here. Uniform Continuity guarantees that for any pair of points \(x,y \in [x_j, x_{j-1}]\), the variation \(|f(x)-f(y)|\) is small. We know the supremum \(M_j\) and the infimum \(m_j\) are just the smallest and largest points \(f\) can take in that interval (Side Note: What if the \(M_j\) and \(m_j\) are not attained?). Hence,

$$ \begin{align} M_j - m_j \leq |f(x) - f(y)| < \epsilon \end{align} $$

Proof

Let \(f\) be a continuous function on \([a,b]\). Since \([a,b]\) is closed and bounded, then \(f\) is uniformly continuous on \([a,b]\). Then for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x - y| < \delta\) and \(x,y \in [a,b]\), then \(|f(x) - f(y)| < \epsilon\). Hence, partition \([a,b]\) into

$$ \begin{align} a = x_0 < x_1 < \ldots < x_n = b, \end{align} $$

and \(x_j - x_{j-1} < \delta\). We then define \(P = \{x_0, x_1, \ldots, x_n\}\). Then by the uniform continuity,

$$ \begin{align} M_j(f) - m_j(f) < \frac{\epsilon}{b-a} \end{align} $$

Hence

$$ \begin{align} U(f,P) - L(f,P) = \sum_{j=1}^{n} (M_j(f)-m_j(f))(x_j - x_{j-1}) < \frac{\epsilon}{b-a} \sum_{j=1}^{n} (x_j - x_{j-1}) = \epsilon. \end{align} $$

References