I have a problem and a solution here please let me know

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Discussion Overview

The discussion revolves around the integrability of the function 1/f, given that f is integrable and bounded below by a positive constant t over the interval [a,b]. Participants explore the implications of these conditions and challenge the reasoning presented in the initial proof.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims that since f is integrable and bounded below by t (where t > 0), then 1/f should also be integrable, arguing that 1/f is bounded and continuous.
  • Another participant questions the conclusion that 1/f is continuous, pointing out that the continuity of f is not established.
  • A different viewpoint suggests that f might not be bounded, proposing to consider the integral in two parts based on the values of f relative to 2t, indicating that the approach may vary depending on whether the integral is Riemann or Lebesgue.
  • One participant notes that the function f(x) = 0 satisfies the initial conditions but leads to 1/f being unbounded, challenging the argument presented.
  • Another participant reminds that one of the conditions is that f(x) > t > 0 for all x in [a,b], which counters the previous example.
  • A participant asks for clarification on the term "bdd," which is later defined as "bounded" by another contributor.

Areas of Agreement / Disagreement

Participants express disagreement regarding the conclusions drawn about the continuity and boundedness of 1/f, with multiple competing views on the implications of the conditions given for f.

Contextual Notes

There are unresolved assumptions regarding the boundedness of f and the implications for the integrability of 1/f, as well as the dependence on the type of integral being considered (Riemann vs. Lebesgue).

shegiggles
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I have a problem here and a solution. Not sure if I am on the right track. Please let me know if this proof is right.

we are given:f :[a,b]-R, f is integrable and bdd below(f is greater
than t(i use t instead of delta) for all x belongs to [a,b] ), t is
greater than 0
claim: 1/f is integrable.
solution: f is integrable implies f is bdd - f is bdd above and below
implies 1/f is bdd below and above - 1/f is bdd-{1}
f is greater than t and t is greater than 0 for all x
- f is greater than 0 for all x. - 1/f can not be infinity for all x-
1/f is continuous.{2}
from {1} & {2} 1/f is a bdd continuous function .
hence 1/f is integrable.
 
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How can you conclude that 1/f is continuous when you don't even know that f is?
 
f might not be bounded. I would consider the integral in two parts: One where f<2t and the other where f>= 2t (say). That is one integral where f is big and one where f is not so big.

As far as how to split up the interval that way would depend on the kind of integral you are doing. Lebesgue - no problem: If you are only dealing with Riemann - may have to turn to (piecewise) continuity to separate the interval accordingly.
 
f(x) identically equal to 0 satisfies all the conditions you give but 1/f certainly is not bounded!
 
Halls, one of the conditions was f(x)>t>0 for all x in [a,b].
 
what is bdd?
 
murshid islam: bdd= bounded.

matt grime- I really do need to learn to read, don't I!
 

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