Integrability of f^2: Show f is Integrable

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SUMMARY

The discussion centers on proving that if the function f is positive and bounded over the interval [a,b], and if f^2 is integrable over [a,b], then f must also be integrable over the same interval. Participants explore the relationship between the upper and lower sums of f^2 and how to derive bounds for f based on these sums. Key insights include the use of inequalities involving the minimum value m of f and the behavior of f as it approaches zero, emphasizing the importance of the bounded nature of f in the integrability proof.

PREREQUISITES
  • Understanding of Riemann integrability
  • Familiarity with upper and lower sums in integration
  • Knowledge of properties of bounded functions
  • Basic algebraic manipulation of inequalities
NEXT STEPS
  • Study the properties of Riemann integrable functions
  • Learn about the relationship between integrability and boundedness
  • Explore the concept of partitions in Riemann integration
  • Investigate the implications of the Mean Value Theorem for integrals
USEFUL FOR

This discussion is beneficial for students of calculus, particularly those studying real analysis, as well as educators and anyone looking to deepen their understanding of integrability conditions for functions.

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Homework Statement



Let f be positive and bounded over [a,b]. If f^2 is integrable over [a,b], then show that f is as well.

The Attempt at a Solution



I'm just trying to use the fact that the upper and lower sums of f^2 over a partition P are arbitrarily close, and then somehow find an upper bound for the difference of the upper and lower sums of f based on that. I've tried separating it into cases, where the max and min of f on an interval are >=1, <1, etc. but it hasn't really led anywhere. If anyone could give me some sort of hint, i'd really appreciate it =)
 
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It's fairly easy if you know f(x)>=m for m>0 isn't it? Since f(x1)^2-f(x2)^2=(f(x1)-f(x2))*(f(x1)+f(x2)) and you know f(x1)+f(x2)>=2m. So f(x1)^2-f(x2)^2>=(f(x1)-f(x2))*2m. Now let m approach zero. Why can you ignore the part of the sums coming from f(x)<m? Remember the interval of integration is [a,b]. It's finite.
 
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