I have an Analysis problem and a solution. please help me

[shegiggles]

I have this problem here and I have a solution for it. I am not really sure if I am right on this. Please let me know if this proof is right.

Thanks

we are given:f :[a,b]-R, f is integrable and bounded 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.

 

[stapel]

Possibly part of the confusion stems from the presention. Would it be possible to have your instructor provide a version with capitalization, punctuation, and spelled-out words? Also, what is the meaning of the various "minus" signs sprinkled throughout?

Thank you.

Eliz.

 

[pka]

Also note that you are not given that the function f is continuous. How do you know that 1/f is continuous?

However, you do know that for any \(\displaystyle \varepsilon > 0\) there is a partition of [a,b] over which \(\displaystyle \overline {\int f } - \underline {\int f } < \varepsilon\).

How does a lower sum of f compare to an upper sum of 1/f on the same subinterval?