# do i use integration?

• Jun 3rd 2006, 09:37 AM
Nichelle14
do 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 PM
ThePerfectHacker
Quote:

the condition |f(a) -f(b)| is less than
|a-b| for all a, b on [0,1]
Does this not show that $\displaystyle f$ is countinous of $\displaystyle [0,1]$? Now use the integration theorem which states that every closed countinous function is Riemann integrable.
• Jun 4th 2006, 08:27 AM
Nichelle14
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 AM
TD!
Bounded, continuous function are Riemann integrable, haven't you seen that?
• Jun 4th 2006, 09:53 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
Bounded, continuous functions

Perhaps, the use of "bounded" was unnesseray because a countinous function on a closed interval is bounded.
• Jun 4th 2006, 09:55 AM
TD!
Then I would've needed to include 'on a closed interval' :)
• Jun 4th 2006, 10:22 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
Then I would've needed to include 'on a closed interval' :)

Good point.

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