Prove Riemann Integrability of Function f on [0,1]

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SUMMARY

The function f defined on the interval [0,1] is Riemann integrable. It is characterized by f(x) = 0 for irrational x and f(x) = 1/q for rational x expressed in lowest terms p/q. The proof utilizes the properties of upper and lower sums, demonstrating that the upper sum converges to 0 while the lower sum remains bounded, thus satisfying the criteria for Riemann integrability. The conclusion confirms that the integral of f over [0,1] is equal to 0.

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  • Understanding of Riemann integrability criteria
  • Familiarity with rational and irrational numbers
  • Knowledge of upper and lower sums in integration
  • Basic concepts of limits and convergence
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Given a function [tex]f: [0,1] \to \mathbb{R}[/tex]. Suppose [tex]f(x) = 0[/tex] if x is irrational and f(x) = 1/q if x = p/q, where p and q are relatively prime.
Prove that f is Riemann integrable.
 
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