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Friday, February 3, 2023

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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