2
$\begingroup$

Why are functions that are continuous over $[a,b]$ integrable over $[a,b]$?

Why is it that to be Riemann-integrable the infimum of the upper sums and the supremum of the lower sums have to be equal?

Can someone explain me this in an easy way?

$\endgroup$
  • 5
    $\begingroup$ The second question is not really a "why" question. That's just the definition. $\endgroup$ – Matt Samuel Jun 21 '16 at 14:18
  • 1
    $\begingroup$ This is a duplicate, but the easiest argument approach is to use the fact that continuous functions on compact intervals are uniformly continuous. $\endgroup$ – symplectomorphic Jun 21 '16 at 14:27
  • $\begingroup$ Can you give the title of your textbook ? $\endgroup$ – Tony Piccolo Jun 22 '16 at 22:13
  • $\begingroup$ @TonyPiccolo why? $\endgroup$ – Alena Jun 22 '16 at 22:13
  • 2
    $\begingroup$ I wonder why you asked the second question: perhaps your textbook defines Riemann integral as a limit ... $\endgroup$ – Tony Piccolo Jun 22 '16 at 22:17
0
$\begingroup$

The answer to your second question is "that's the definition of Riemann-integrable - the common value is the Riemann integral."

The definition makes intuitive sense if you think about the pictures for the upper and lower sums and want to study functions for which those sums approximate the area.

The answer to your first question is a theorem that starts with the definition of continuity and proves the inf and sup of the upper and lower sums are the same. The proof depends on the intuitive idea that values of a continuous function at close together points are close together.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.