**Theorem: **

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

**Proof:**

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

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Friday, February 3, 2023

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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