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?
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5The second question is not really a "why" question. That's just the definition. – Matt Samuel Jun 21 '16 at 14:18
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1This is a duplicate, but the easiest argument approach is to use the fact that continuous functions on compact intervals are uniformly continuous. – symplectomorphic Jun 21 '16 at 14:27
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Can you give the title of your textbook ? – Tony Piccolo Jun 22 '16 at 22:13
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@TonyPiccolo why? – Alen Jun 22 '16 at 22:13
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2I wonder why you asked the second question: perhaps your textbook defines Riemann integral as a limit ... – Tony Piccolo Jun 22 '16 at 22:17
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hm there is also this definition @TonyPiccolo – Alen Jun 22 '16 at 22:28
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The point is this. If you start from the definition of integral as a limit, your question is well-founded. Otherwise, if you start from the definition a la Darboux (upper and lower sums), you asked essentially why A is A (banal of course). – Tony Piccolo Jun 22 '16 at 22:32
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Voting to close. Several comments asking for explanations to help eliminate apparent absurd aspects of the question were met only by stonewalling from the OP. In the end only remains a bad question. – Did Jun 26 '16 at 14:22
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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.
Ethan Bolker
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