Friday, February 3, 2023
HomeMathematicsCalculusEvery continuous function f defined on () 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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