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Every monotonic function on [a, b] is Riemann integrable

Theorem

Every monotonic function on [a, b] is Riemann integrable.

Proof:

Let ​\( f:[a, b]\to \mathbb{R} \)​ be monotonic function.
If f is constant function then obviously it is Riemann integrable.
WLOG, suppose f is strictly increasing.

\[ \therefore f(a)<f(x_1)<f(x_2)<f(b), a<x_1<x_2<b\]

\( \therefore f(b)-f(a)>0 \)
Choose a partition of [a, b] for ​\( \epsilon>0 \)​ such that,
\( P=\{a=x_0, x_1, x_2, … , x_n=b\} \)​ with ​\( \|P\|<\frac{\epsilon}{f(b)-f(a)} \)
for ​\( [x_{k-1}, x_k], \|x_k-x_{k-1}\|<\frac{\epsilon}{f(b)-f(a)} \)
Consider,
U(f, P)-L(f, P)
\( =\displaystyle\sum_{k=1}^{n}(M_k-m_k)\Delta_{x_k} \)
\( =\displaystyle\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})]\Delta_{x_k} \)
\( \leq\displaystyle\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})]\|P\| \)
\( <\displaystyle\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})]\frac{\epsilon}{f(b)-f(a)} \)
\( =\frac{\epsilon}{f(b)-f(a)}\displaystyle\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})] \)
\( =\frac{\epsilon}{f(b)-f(a)}\times[f(b)-f(a)] \)
\( =\epsilon \)
\( \therefore U(f, P)-L(f, P)<\epsilon \)​ 
\( \therefore \)​ by Riemann criterion,
f is Riemann integrable.
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