Riemann Integral - little proof help

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SUMMARY

The discussion focuses on proving the integrability of the function 1/f given that f is integrable on the interval [a,b] and f(x) > C for some constant C. The key approach involves establishing that the lower and upper sums of 1/f can be made arbitrarily close by manipulating the bounds of f, specifically using the inequalities 1/M ≤ 1/f(x) ≤ 1/m. The proof hinges on the relationship between the bounds of f and the constant C, leading to a method for bounding the difference between the upper and lower sums of 1/f.

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Homework Statement



Suppose f is integrable for all x in[a,b] and f(x)>C ( C is some constant),
Must show that 1/f is also integrable.


Homework Equations



f is integrable implies Upf-Lpf<\epsilon for some partition in [a,b]



The Attempt at a Solution



Therefore, I must come up with a good \epsilon such that
Lp(1/f) - Up(1/f) <\epsilon

Also f is bounded because it's integrable so there must be some m,M such that
f([a,b])= [m,M]
in other words f acheives it's minimum and maximum points on the interval.
 
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Some hints. For simplicity let's just look at one subinterval I=[x_i,x_{i+1}]

On I, we have m\le f(x)\le M (although really there is no guarantee that m and M are achieved, we still know f is bounded as you said).

Where does C fit in this inequality?

Take reciprocals of the inequalities 1/M\le 1/f(x)\le 1/m and put 1/c in the correct place, and furthermore where is 0?

Suppose M-m<delta.

Then \frac1m-\frac1M=\frac{M-m}{Mm} and you can find an upper bound for the last fraction in terms of delta and C.
 

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