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.
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.the condition |f(a) -f(b)| is less than
|a-b| for all a, b on [0,1]
Good point.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.