# Theorem :

### A bounded function $f:[a, b]\to \mathbb{R}$ is Riemann integrable iff for every $\epsilon>0$ there exist a partition $P_\epsilon$ of [a, b] such that $U(f, P_\epsilon)-L(f, P_\epsilon)<\epsilon$.

# Proof :

Suppose f is Riemann integrable on [a, b].

Let $\epsilon>0$ be arbitrary.

$\int\limits_\underline{a}^bf(x)dx=\int\limits_a^\underline{b}f(x)dx$ ..... (1)

By definition of lower integral,

$\int\limits_\underline{a}^bf(x)dx=sup\{L(P, f)$, P is partition of $[a, b]\}$

By definition of supremum,

$\exists$ a partition $P_1$ of [a, b] such that,

$\int\limits_\underline{a}^bfdx-\frac{\epsilon}{2}<L(P_1, f)\leq\int\limits_\underline{a}^bfdx$ ..... (2)

By definition of upper integral,

$\int\limits_a^\underline{b}f(x)dx=inf\{U(P, f)$, P is a partition of $[a, b]\}$

By definition of infimum,

$\exists$ some partition $P_2$ of [a, b] such that,

$\int\limits_a^\underline{b}fdx\leq U(P_2, f)<\int\limits_a^\underline{b}fdx+\frac{\epsilon}{2}$ ..... (3)

Let $P_\epsilon=P_1\cup P_2$ be the refinement of $P_1$ and $P_2$.

$U(P_\epsilon, f)\leq U(P_2, f)$

$L(P_1, f)\leq L(P_\epsilon, f)$

Consider,

$U(P_\epsilon, f)-L(P_\epsilon, f)$

$\leq\int\limits_a^\underline{b}f(x)dx+\frac{\epsilon}{2}-\int\limits_\underline{a}^bf(x)dx+\frac{\epsilon}{2}$ .... from (1), (2) & (3)

$=\epsilon$

Hence, we have partition $P_\epsilon$ such that,

$U(P_\epsilon, f)-L(P_\epsilon, f)<\epsilon$

Conversely,

Let $\epsilon>0$ be arbitrary and for this $\epsilon$,

$\exists$ a partition $P_\epsilon$ such that, $U(P_\epsilon, f)-L(P_\epsilon, f)<\epsilon$.

Now, $U(P_\epsilon, f)\geq L(P_\epsilon, f)$

$\implies 0\leq U(P_\epsilon, f)-L(P_\epsilon, f)<\epsilon$

Since, this inequality true for every $\epsilon>0$,

$\therefore\int\limits_\underline{a}^{b}fdx=\int\limits_a^\underline{b}fdx$

Hence, f is Riemann integrable.

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