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Theorem:
Every continuous function f defined on [a, b] is Riemann integrable
Proof:
Since, every function defined on closed and bounded interval is uniformly continuous function on [a, b],
​\( \therefore \)​ By definition of uniformly continuous function,
​\( \forall \epsilon>0, \exists \delta>0 \)​ (​\( \delta \)​ depends only on ​\( \epsilon \)​) such that,
\[ \|x-y\|<\delta\implies\|f(x)-f(y)\|<\frac{\epsilon}{b-a},\forall x, y\in[a, b] \]
​ ……………. (1)
Let ​\( P=\{a=x_0, x_1, … ,x_n=b\} \)​ be a partition of [a, b] such that ​\( \|P\|<\delta \)​.
Since, f is continuous on [a, b], it is continuous on ​\( [x_{k-1}, x_k] \)​.
Also, f is continuous on closed and bounded interval then f attains its bounds.
Let ​\( y_k, z_k\in [x_{k-1}, x_k] \)​ such that,
​\( m_k=f(y_k) \)​ and ​\( M_k=f(z_k) \)​, k=1,2, … , nÂ
​\( \|y_k-z_k\|=\leq\|x_k-x_{k-1}\|\leq\|P\|<\delta \)​
​\( \|f(y_k)-f(z_k)\|<\frac{\epsilon}{b-a} \)​ ….. from (1)
i.e. ​\( \|m_k-M_k\|<\frac{\epsilon}{b-a} \)​
i.e. ​\( \|M_k-m_k\|<\frac{\epsilon}{b-a} \)​
​\( \therefore M_k-m_k<\frac{\epsilon}{b-a} \)​ (​\( \because M_k\geq m_k \)​) ….. (2)
Consider,
U(f, P)-L(f, P)Â
​\( =\displaystyle\sum_{k=1}^{n}M_k.\Delta_{x_k}-\displaystyle\sum_{k=1}^{n}m_k.\Delta_{x_k} \)​
​\( =\displaystyle\sum_{k=1}^{n}(M_k-m_k)\Delta_{x_k} \)​
​\( <\displaystyle\sum_{k=1}^{n}\Big(\frac{\epsilon}{b-a}\Big)(x_k-x_{k-1}) \)​
​\( =\Big(\frac{\epsilon}{b-a}\Big)\displaystyle\sum_{k=1}^{n}(x_k-x_{k-1}) \)​
​\( =\frac{\epsilon}{(b-a)}(b-a) \)​
​\( =\epsilon \)​
​\( \therefore U(f, P)-L(f, P)<\epsilon \)​
​\( \therefore \)​ for ​\( \epsilon>0 \)​, ​\( \exists \)​ a partition P such that,
​\( U(f, P)-L(f, P)<\epsilon \)​Â
​\( \therefore \)​ by Riemann criterion,
f is Riemann integrable.
Also Read:Â First and Second fundamental theorem of calculus
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