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

[shegiggles]

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.

 

[DeadWolfe]

How can you conclude that 1/f is continuous when you don't even know that f is?

 

[gammamcc]

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.

 

[HallsofIvy]

f(x) identically equal to 0 satisfies all the conditions you give but 1/f certainly is not bounded!

 

[matt grime]

Halls, one of the conditions was f(x)>t>0 for all x in [a,b].

 

[murshid_islam]

what is bdd?

 

[HallsofIvy]

murshid islam: bdd= bounded.

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