Show that the following function is Riemann integrable.

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SUMMARY

The function f: [0,1] -> R defined by f(x) = 1 if x = 1/k for some k, and f(x) = 0 otherwise, is Riemann integrable on the interval [0,1]. The key to proving this is demonstrating that the lower and upper Riemann sums converge to the same limit. By applying the definition of Riemann integrability, one must show that for any ε > 0, a suitable partition of [0,1] can be constructed such that the difference between the upper sum and the integral is less than ε.

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Homework Statement



Show that the function f: [0,1] -> R defined by:

f(x) = 1, if x=1/k for some k
f(x) = 0, else

is Riemann integrable on [0,1]

Homework Equations





The Attempt at a Solution



I attempted the problem using Cauchy's criterion but found that this function is not Riemann integrable, am i right? or am i going about this the wrong way.
 
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I'm assuming you don't have access to any of the standard equivalent criteria to Riemann integrability. To show that a function is Riemann integrable directly from the definition, you must show that the lower and upper integral are equal, that is, that the lower and upper Riemann sums have a common least upper (greatest lower) bound.

First of all, by looking at the function, you should be able to see immediately what the lower sums should be and therefore what the value of the integral must be. So you just have to prove that the infimum of the upper sums is the right thing. This you do by constructing, for any \varepsilon > 0, a partition of [0,1] such that the upper sum with respect to this partition differs from the value of the integral by less than \varepsilon.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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