21.8 C
Maharashtra
Tuesday, March 28, 2023
HomeMathematicsCalculusEvery monotonic function on is Riemann integrable

Proof:

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

$\therefore f(a)<f(x_1)<f(x_2)<f(b), a<x_1<x_2<b$

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