if f is bounded on [0,1] and satisfies the condition |f(a) -f(b)| is less than

|a-b| for all a, b on [0,1] , prove that f is Riemann integrable on [0,1]

I have always been slightly confused by Riemann. :confused:

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- Jun 3rd 2006, 09:37 AMNichelle14do i use integration?
if f is bounded on [0,1] and satisfies the condition |f(a) -f(b)| is less than

|a-b| for all a, b on [0,1] , prove that f is Riemann integrable on [0,1]

I have always been slightly confused by Riemann. :confused: - Jun 3rd 2006, 06:57 PMThePerfectHackerQuote:

the condition |f(a) -f(b)| is less than

|a-b| for all a, b on [0,1]

- Jun 4th 2006, 08:27 AMNichelle14
Yes. it does show f is continuous. But I don't know how to show that it is Riemann integrable. I don't think I ever grasped that when taught.

Can you explain some more? - Jun 4th 2006, 09:33 AMTD!
Bounded, continuous function are Riemann integrable, haven't you seen that?

- Jun 4th 2006, 09:53 AMThePerfectHackerQuote:

Originally Posted by**TD!**

- Jun 4th 2006, 09:55 AMTD!
Then I would've needed to include 'on a closed interval' :)

- Jun 4th 2006, 10:22 AMThePerfectHackerQuote:

Originally Posted by**TD!**

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What about,

$\displaystyle f(x)=x$ boundend and countinous on $\displaystyle [0,\infty]$ but not riemann integrable. Thus, you need to include 'on a closed interval' anyway.