Absolute value of a function integrable?

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SUMMARY

The discussion centers on proving that if a function f is continuous on the interval (a,b] and the absolute value |f| is bounded on [a,b], then f is integrable on [a,b]. The proof involves using upper and lower bounds to demonstrate that for any ε > 0, a finite subcover from an open cover can be chosen, leading to the conclusion that the upper sum U and lower sum L satisfy U - L < ε. This establishes the integrability of f on the specified interval.

PREREQUISITES
  • Understanding of real analysis concepts, specifically continuity and integrability.
  • Familiarity with the definitions of upper and lower sums in the context of Riemann integration.
  • Knowledge of open covers and compact sets in topology.
  • Proficiency in constructing finite subcovers from open covers.
NEXT STEPS
  • Study the properties of Riemann integrable functions and their criteria.
  • Learn about the concept of compactness in metric spaces.
  • Explore the relationship between continuity and integrability in real analysis.
  • Investigate the use of ε-δ definitions in proving function properties.
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This discussion is beneficial for students and professionals in mathematics, particularly those studying real analysis, as well as educators seeking to clarify the concepts of integrability and continuity in functions.

tomboi03
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this is the question,
Prove that if f is continuous on (a,b] and if |f| is bounded on [a,b] then f is integrable on [a,b]. (note: it is not assumed that f is continuous at a.)

I know you have to use the upper and lower bounds to prove this statement but i don't know where to start?

Thanks
 
Last edited:
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let eps>0
H(x):={y!=x| |f(x)-f(y)|<eps/(b-a)}
Union[H(x)|x in [a,b]]
is an open cover (by continuity of f) of [a,b] a compact set so we may chose a finite subcover
is P is any partition at least as fine as the open cover will have
U-L<eps
qed
 

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