Any continuous function is Riemann-integrable on a closed interval

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The goal of this post is to prove one of the practical foundations of Riemann-integrals.



Theorem

A function that is continuous on \([a,b]\) is integrable on \([a,b]\).  

See also: Riemann integral

Proof. Let \(f(x)\) be continuous on \([a,b]\). By the above theorem, \(f(x)\) is uniformly continuous on \([a,b]\). Therefore, for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that, for all \(x, y\in [a,b]\), \(|x-y| < \delta\) implies \(|f(x) - f(y)| < \frac{\varepsilon}{b - a}\).

Let \(\Delta\) be a partition of \([a,b]\) such that \(a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b\) and its mesh is less than \(\delta\) (i.e., \(x_{i+1} - x_{i} < \delta\) for all \(i= 0, 1, \cdots, n-1\)). Then, for each \(i=0, 1, \cdots, n-1\), if \(x, y \in [x_{i}, x_{i+1}]\), then \(|f(x) - f(y)| < \frac{\varepsilon}{b - a}\). Hence, if we define

\[ \begin{eqnarray} M_i &=& \sup\{f(x) \mid x_i \leq x \leq x_{i+1}\},\\ m_i &=& \inf\{f(x) \mid x_i \leq x \leq x_{i+1}\}, \end{eqnarray} \]

\(M_i - m_i < \frac{\varepsilon}{b - a}\). Therefore, using the upper and lower Riemann sums,

\[ \begin{eqnarray} S_{\Delta} &=& \sum_{i=0}^{n-1}M_i(x_{i+1} - x_{i}),\\ s_{\Delta} &=& \sum_{i=0}^{n-1}m_i(x_{i+1} - x_{i}), \end{eqnarray} \]

we have

\[ \begin{eqnarray*} S_{\Delta} - s_{\Delta} &=& \sum_{i=0}^{n-1}(M_i - m_i)(x_{i+1} - x_{i})\\ &<&\sum_{i=0}^{n-1}\frac{\varepsilon}{b-a}(x_{i+1} - x_{i})\\ &=& \frac{\varepsilon}{b-a}\cdot(b-a) = \varepsilon. \end{eqnarray*} \]

Thus, in particular,

\[ \inf_{\Delta}S_{\Delta} - \sup_{\Delta}s_{\Delta} < \varepsilon. \]

However, since \(\varepsilon > 0\) is arbitrary, it follows that

\[ \inf_{\Delta}S_{\Delta} = \sup_{\Delta}s_{\Delta}. \]

Therefore, \(f(x)\) is Riemann-integrable. ■




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