Showing that a discontinuous function is integrable

  • Thread starter Faris
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  • #1
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Hi,
First, I want to thank you people for the help.

1. Define a function f by setting f(x)=1 if x=1/m for some positive integer m (e.g. f(1)=1, f(1/2)=1, f(1/3)=1,...) and f(x)=0 otherwise. Show that f is integrable on the interval [0,1] by proving that the limit of the lower Riemann sums for the regular partitions P_n equals the limit of the corresponding upper Riemann sums.



2. Lower Riemann sum = L(f,P_n)
Upper Riemann sum = U(f,P_n)




3. I've tried the following argument, but I don't think it's enough for a proof:

L(f,P_n) = 0 because by taking n to infinity, the minimum values for f(x) in the delta x's are zeros, which makes L(f,P_n)=0.

U(f,P_n), however, is the problem here.


Thanks in advance.
 

Answers and Replies

  • #2
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Hint: If you have a partition [itex]P_n[/itex], then each piece of a partition that contains a point where f is nonzero has an area of [itex]\frac{1}{n}[/itex]. What you need to show is that for any [itex]\epsilon[/itex], there is an [itex]n[/itex] such that no more than [itex]n\epsilon[/itex] of the pieces contain a point where f is nonzero.

Try to do this example and you'll probably get the idea of the proof:

Exercise:
Let [itex]\epsilon=0.1[/itex]. Show that you can make [itex]U,P_{1000} \le \epsilon[/itex] by covering every value 1/n (for positive integers n) by selecting 100 intervals from [itex]P_{1000}[/itex].
 

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