# 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.