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

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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