Lemma 0.9
If a function \(f\) is Reimann integrable on \([a,b]\), then \(f\) is bounded on \([a,b]\). (i.e. there exists an \(M\) such that \(|f(x)| \leq M\) for all \(x \in [a,b]\).
Proof
Suppose that \(f\) is unbounded and suppose that \(f\) is Riemann integrable on \([a,b]\). Then by definition, there exists a \(\mathcal{P}_{\epsilon}\) such that for any refinement \(\mathcal{P} \supset \mathcal{P}_{\epsilon}\) and any choice of tags, the Riemann sum for with respect to the set of tags must be close to the integral. i.e. if we set \(\epsilon_0 = 1\), then
$$
\begin{align}
|S(f, \mathcal{P}, t_j) - I(f)| < \epsilon_0 = 1.
\end{align}
$$
By the reverse triangle inequality, this implies that
$$
\begin{align}
|S(f, \mathcal{P}, t_j)| - |I(f)| &\leq |S(f, \mathcal{P}, t_j) - I(f)| < 1
\end{align}
$$
Hence,
$$
\begin{align}
|S(f, \mathcal{P}, t_j) &< 1 + |I(f)| \quad \quad \quad (1)
\end{align}
$$
In other words, no Riemann sum is allowed to be extremely large or father away from \(|I(f)|\). Now, take any partition
$$
\begin{align}
\mathcal{P} = \{x_0,\ldots,x_n\}
\end{align}
$$
Since \(f\) is unbounded on \([a,b]\), we must be able to find at least one interval \([x_{i-1}, x_i]\) on which \(f\) is unbounded. By definition, this means for any \(K > 0\), there exists some point \(t_i \in [x_{i-1}, x_i]\) such that
$$
\begin{align}
|f(t_i)| > K
\end{align}
$$
Fix all the tags \(t_j\) where \(j \neq i\). Let
$$
\begin{align}
C = \sum_{j \neq i}^n f(t_j)(x_j - x_{j-1})
\end{align}
$$
We know that the Riemann sum is defined to be
$$
\begin{align}
S = C + f(t_i)(x_i - x_{i-1})
\end{align}
$$
The goal is have \(|f(t_i)|\) (where \(f\) is unbounded) be as large as we can so that the Riemann sum of the partition \(S\) exceed the bound we derived in (1). That is
$$
\begin{align}
|S| = |C + f(t_i)(x_i - x_{i-1})| \geq 1 + |I(f)|
\end{align}
$$
Hence, let
$$
\begin{align}
K = \frac{1 + |I(f)| + |C|}{x_i - x_{i-1}}
\end{align}
$$
Because \(f\) is unbounded on \([x_i, x_{i-1}]\), there exists a tag \(t_i\) such that
$$
\begin{align}
|f(t_i)| &> K = \frac{1 + |I(f)| + |C|}{x_i - x_{i-1}} \\
\end{align}
$$
so
$$
\begin{align}
|f(t_i)|(x_i - x_{i-1}) &> 1 + |I(f)| + |C| \quad\quad\quad (2)
\end{align}
$$
But now observe that the Riemann Sum is
$$
\begin{align}
S = C + f(t_i)(x_i - x_{i-1})
\end{align}
$$
Using the reverse triangle inequality with \(A = f(t_i)(x_i - x_{i-1})\) and \(B = C\), we know that \(||A| - |-B|| \leq |A-(-B) = A+B|\). But \(A+B = S\). Hence
$$
\begin{align}
|S| \geq \left||f(t_i)(x_i - x_{i-1})| - |C|\right|
\end{align}
$$
Using (2),
$$
\begin{align}
|f(t_i)|(x_i - x_{i-1}) - |C| > (1 + |I(f)| + |C|) - |C| = 1 + |I(f)|.
\end{align}
$$
Thus
$$
\begin{align}
|S| > 1 + |I(f)|
\end{align}
$$
This is a contradiction to (1). Hence, \(f\) must be bounded. \(\ \blacksquare\)
References