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October 29th, 2010, 08:05 PM   #1
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Prove Lower Integral <= 0 <= Upper Integral

Suppose f:[a, b]-> R is bounded function
f(x)=0 for each rational number x in [a, b]
Prove Lower Integral <= 0 <= Upper Integral

Proof:
f(x) = 0 when x is rational
both L(f, p) = U(f, P) = 0
and L(f, p) <= Lower Integral <= Upper Integral <= U(f, p)

This function seems like discontinous even though there aren't any information of functional value when x is NOT rational. It looks like that the Intermediate Value Theorem needs to be appplied.

So I have to prove that the Lower Integral <=0, and the Upper Integral >=0. So the function itself has to cross f(x)=0 isn't it?

Any suggestions would be greatly appreciated.
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October 29th, 2010, 08:27 PM   #2
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Re: Prove Lower Integral <= 0 <= Upper Integral

Never mind, I got it.
Every interval of nonzero size contains a rational number.
So the min of f(x) on the interval MUST be <=0
and the max of f(x) on the interval is >=0.
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