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Friday, June 9, 2023
HomeMathematicsCalculusEvery continuous function f defined on () is Riemann integrable

Proof:

Since, every function defined on closed and bounded interval is uniformly continuous function on [a, b],
â€‹$$\therefore$$â€‹Â By definition of uniformly continuous function,
â€‹$$\forall \epsilon>0, \exists \delta>0$$â€‹ (â€‹$$\delta$$â€‹ depends only on â€‹$$\epsilon$$â€‹) such that,

$\|x-y\|<\delta\implies\|f(x)-f(y)\|<\frac{\epsilon}{b-a},\forall x, y\in[a, b]$

â€‹ ……………. (1)

Let â€‹$$P=\{a=x_0, x_1, … ,x_n=b\}$$â€‹ be a partition of [a, b] such that â€‹$$\|P\|<\delta$$â€‹.
Since, f is continuous on [a, b], it is continuous on â€‹$$[x_{k-1}, x_k]$$â€‹.
Also, f is continuous on closed and bounded interval then f attains its bounds.
Let â€‹$$y_k, z_k\in [x_{k-1}, x_k]$$â€‹Â such that,
â€‹$$m_k=f(y_k)$$â€‹ and â€‹$$M_k=f(z_k)$$â€‹, k=1,2, … , nÂ
â€‹$$\|y_k-z_k\|=\leq\|x_k-x_{k-1}\|\leq\|P\|<\delta$$â€‹
â€‹$$\|f(y_k)-f(z_k)\|<\frac{\epsilon}{b-a}$$â€‹Â ….. from (1)
i.e. â€‹$$\|m_k-M_k\|<\frac{\epsilon}{b-a}$$â€‹
i.e. â€‹$$\|M_k-m_k\|<\frac{\epsilon}{b-a}$$â€‹
â€‹$$\therefore M_k-m_k<\frac{\epsilon}{b-a}$$â€‹Â  (â€‹$$\because M_k\geq m_k$$â€‹) ….. (2)
Consider,
U(f, P)-L(f, P)Â
â€‹$$=\displaystyle\sum_{k=1}^{n}M_k.\Delta_{x_k}-\displaystyle\sum_{k=1}^{n}m_k.\Delta_{x_k}$$â€‹
â€‹$$=\displaystyle\sum_{k=1}^{n}(M_k-m_k)\Delta_{x_k}$$â€‹
â€‹$$<\displaystyle\sum_{k=1}^{n}\Big(\frac{\epsilon}{b-a}\Big)(x_k-x_{k-1})$$â€‹
â€‹$$=\Big(\frac{\epsilon}{b-a}\Big)\displaystyle\sum_{k=1}^{n}(x_k-x_{k-1})$$â€‹
â€‹$$=\frac{\epsilon}{(b-a)}(b-a)$$â€‹
â€‹$$=\epsilon$$â€‹
â€‹$$\therefore U(f, P)-L(f, P)<\epsilon$$â€‹
â€‹$$\therefore$$â€‹ for â€‹$$\epsilon>0$$â€‹, â€‹$$\exists$$â€‹ a partition P such that,
â€‹$$U(f, P)-L(f, P)<\epsilon$$â€‹Â
â€‹$$\therefore$$â€‹Â by Riemann criterion,
f is Riemann integrable.
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