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.
Also Read: First and Second fundamental theorem of calculus
