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Mathematics
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WITH WHAT?
integrals and reimann sums
A function does not have to be continuous in order for it to be Riemann integrable. However, the one-way implication is true: every continuous real-valued function on a closed interval [a,b] is Riemann integrable there. The converse to this is false. However, I did not (I hope!) claim that Riemann integrable functions must be continuous. - Whenever f is a bounded real-valued function on [a,b], then you can define the Riemann upper sums and the Riemann lower sums for f. This does not work if the function f is unbounded, though. For example, if the function f is, say, 1/x for x not equal to 0, but f(0)=0, then you can not define Riemann lower/upper sums for f on [-1,1]. - Even when f is a bounded, real-valued function on [a,b], you can have problems. For example, if the function f(x) is defined to be 1 when x is rational but 0 when x is irrational, and you try to find the Riemann integral on the interval [0,2], say, then all of the Riemann upper sums come out to be 2, while the Riemann lower sums are all 0. As a result, the Riemann upper integral is 2, and the Riemann lower integral is 0. Since these are different, the Riemann integral does not exist here, and f is not Riemann integrable on [0,2].

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