# 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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