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HomeMathematicsCalculusEvery continuous function f defined on () is Riemann integrable

Every continuous function f defined on \([a, b]\) is Riemann integrable

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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, … , n 
\( \|y_k-z_k\|=\leq\|x_k-x_{k-1}\|\leq\|P\|<\delta \)
\( \|f(y_k)-f(z_k)\|<\frac{\epsilon}{b-a} \)​ ….. 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) 
\( =\displaystyle\sum_{k=1}^{n}M_k.\Delta_{x_k}-\displaystyle\sum_{k=1}^{n}m_k.\Delta_{x_k} \)
\( =\displaystyle\sum_{k=1}^{n}(M_k-m_k)\Delta_{x_k} \)
\( <\displaystyle\sum_{k=1}^{n}\Big(\frac{\epsilon}{b-a}\Big)(x_k-x_{k-1}) \)
\( =\Big(\frac{\epsilon}{b-a}\Big)\displaystyle\sum_{k=1}^{n}(x_k-x_{k-1}) \)
\( =\frac{\epsilon}{(b-a)}(b-a) \)
\( =\epsilon \)
\( \therefore U(f, P)-L(f, P)<\epsilon \)
\( \therefore \)​ for ​\( \epsilon>0 \)​, ​\( \exists \)​ a partition P such that,
\( U(f, P)-L(f, P)<\epsilon \)​ 
\( \therefore \)​ by Riemann criterion,
f is Riemann integrable.
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