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HomeMathematicsCalculusA bounded function ​( f:to mathbb{R} )​ is Riemann integrable iff for...

# 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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